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THE 

GRAPHIC METHOD 

BY 

INFLUENCE LINES 

FOR 

BRIDGE AND ROOF COMPUTATIONS 



BY 

WILLIAM H. BURR, C.E. 

^o/essor of Civil Engineering in Columbia University in the City of Neiv York 

Member of the A jnerican Society of Civil Engineers ; Member of the 

1 nstitution of Civil Engineers of Great Britain 



MYRON S. FALK, Ph.D. 

hecturer in Cii'tl Engineering in Colutnhia Uni-versity in the City of Neiv York 
Associate Member of the American Society o/ Civil Engineers 



SECOND EDITION 
FIRST THOUSAND 



NEW YORK 

JOHN WILEY & SONS 

London: CHAPMAN & HALL, Limited 
1908 



c. 



-^^•^•, 



TuiRARY of CONGiRESS 
I two Copies Keceivea 

FEB 15 1908 

' (JOWfllSii'. !:.iHf)' _ 
CLASS 4 '<''^<:' **"■ 

I qq 135' 

COPY B. 



Copyright, 1905, 1908, 

BY 

WM. H. BURR 

AND 

MYRON S. FALK 



ROBERT DRUMMONI>, PRINTER, NEW YORK 



«- [p0'7(^ 



PREFACE. 



This book exhibits an entirely modem graphical treat- 
ment, by the method of influence lines, of simple statically 
determinate structures such as bridges and roof-trusses, 
three-hinged arches, cantilevers, and other constructions 
of the same general class. The simplicity, elegance, and 
generality of procedures by influence lines and areas afford 
quick and eminently practical methods of computing 
stresses in all forms of trusses, whether with straight, 
curved, or broken outlines and with any single or multiple 
systems of bracing, whether square or skew, or with any 
manner of loading whatever. The ease and clearness 
with which the greatest stresses, both main and counter, in 
web and chord m_embers, may be determined for any 
specified system of loading has already brought the treat- 
ment into general use in practical structural design. Taken 
as a whole the method commends itself not only as a 
simple universal system of analysis, but also as a remark- 
ably labor-saving means of making computations in struc- 
tural work. Some knowledge of elementary algebraic 
methods, including the method of moments or sections, 
is required, but nothing more. 

The graphical method of finding the deflections of all 
points of trusses and other structures, given in Chapter V, 
has also marked advantages of extreme simplicity, gen- 
erality, and saving of labor, when compared with the alge- 



IV PREFACE. 

braic method. The effect of the strain in every member 
of the structure has its proper place and value in the final 
result, thus securing complete accuracy. 

The reader of the book will find that the methods of 
treatment set forth in it cover plate girders and solid 
beams, as well as articulated or jointed structures like 
ordinary bridge- and roof -trusses. The stresses in roof- 
trusses caused by the flexure of the supporting columns 
are treated in detail. 

Sufficient problems are introduced to illustrate clearly 
the practical applications of the various methods of treat- 
ment, but it should always be borne in mind that the 
greatest advantage will accrue to those students who work 
out original problems and make independent practical 
applications. 

The same general observations hold in connection with 
the complete design of the members of a railroad bridge 
given at the end of the book, to which, in the determination 
of the stresses, the previously deduced methods are applied. 

The application of these lines to statically indeter- 
minate structures, such as two-hinged and fixed ended 
arches, swing bridges, suspension bridges, and certain 
types of cantilevers, will appear in another volume. 

W. H. B. 
M. S. F. 

Columbia University, February 2, 1905. 



PREFACE TO SECOND EDITION. 



In issuing the second edition of this book, use is made 
of the opportunity to correct such typographical errors as 
inevitably are found in the diagrams and numerical work 
necessarily employed in the preparation of such a work. 
All such corrections have been made so far as they have 
been discovered, and it is believed that no material inaccu- 
racies of that character remain. 

The method of determining deflections of trusses has 
been extended to include those with subdivided panels. 
It is believed that as this type of truss is much used at 
present the new material will be found of practical value. 

The largest addition of new matter consists of the 
eighteen pages devoted to suspended cantilevers. With 
the extension of bridge work into the field of long spans, 
the treatment of this type of structure adds materially to 
the scope and value of the book. 

W. H. B. 
M. S. F. 

Columbia University, January 8, 1908. 

V 



CONTENTS. 



CHAPTER I. 

GENERAL PRINCIPLES OF GRAPHIC STATICS, AND APPLICATION 
TO ROOF-TRUSSES. 

ART. PAGB 

1. Definitions i 

2. Resultant of any Number of Forces 3 

3. Application of Force Polygon to Finding of Unknown Quan- 

tities ." 4 

4. System of Notation 5 

Example 6 

5. Determination of Kind of Stresses 7 

6. Crane-trusses 8 

7. Non-concurrent Coplanar Forces, and the Funicular Poly- 

gon 13 

8. Some Special Features of the Funicular Polygon 17 

9. Lines of Action of all Forces Known, but Magnitudes and 

Directions of Two Forces Unknown 20 

10. Moments in a Beam 23 

11. All Forces Given Except Two, of which the Line of Action 

OF One and the Point of Application of the Other are 

Known 25 

12. All Forces Given Except Three, with the Lines of Action 

OF those Three Known 26 

13. Stresses in Roof-trusses 29 

Preliminary 29 

Snow Load 30 

Wind Load 31 

Roof Covering 32 

14. Stresses in a Roof-truss, Both Ends Fastened 33 

15. Stresses in a Roof-truss, One End on Rollers 38 

Counterbraces. 4a 

vii 



viii CONTENTS. 

ART. PAGB 

i6. Fink Roop-truss . . . 42 

17. Unsymmetrical Trusses 46 

18. Bending of Supporting Columns of Roofs 48 

Case I. Posts Hinged at Top and Base 50 

Case U. Posts Hinged at Top and- Fixed at Base 53 



CHAPTER II. 
INFLUENCE LINES FOR SIMPLY SUPPORTED BRIDGE-TRUSSES. 

1. Influence Line for Reaction 60 

2. Influence Line for Shear 61 

3. Influence Line for the Reactions of a Series of Concentrated 

Loads 63 

4. Influence Line for Maximum Shear, for a Series of Concen- 

trated Loads 66 

5. Influence Line for Moments 67 

6. Criterion for Maximum Moment at any Section of a Beam 70 

7. Maximum Moments in a Beam 71 

8. Maximum Stresses in the Web Members of a Truss with Paral- 

lel AKTD Horizontal Chords 74 

9. Maximum Stresses in the Chord Members of a Truss with 

Parallel Chords 77 

10. Influence Lines between Adjacent Panel Points] 78 

11. Moment Influence Lines for any Truss 79 

12. Variation of Moment within a Panel for a Fixed Position 

OF THE Loading 83 

13. Problem in Finding the Maximum Stress in the Loaded Chord 

Member of a Truss with Web Members all Inclined 84 

14. Determination of Stresses in Three Non-concurrent Mem- 

bers OF A Truss 87 

15. Stresses in the Web Members of any Simply Supported Truss. 88 

16. Influence Line for Stress in any Web Member of a Simply 

Supported Truss 91 

Method 1 91 

Method II 94 

17. Criterion to Determine Position of Loading for Maximum 

Web Stress 97 

Application of Critenon to a Prohcem 99 

18. Trusses with Subdivided Panels 103 

■ 19. Maximum Web Stresses in Trusses with Subdivided Panels. . . . 105 

20. Maximum Chord Stresses in Trusses with Subdivided Panels . . 106 

Unloaded Chord 106 

Loaded Chord 109 



CONTENTS. IX 

ART. PAGE 

21. Counter-stresses in a Vertical Post at an Angle in a Chord. . no 

Application to an Example 115 

22. Influence Line por Stress in the Web Member of a Truss 

when the Centre of Moments Falls within the Limits of 
THE Truss 117 

23. Influence Lines for Skew Bridges 119 

24. Influence Lines for Double-intersection Trusses 122 

Chord Members 123 

Web Members 125 

CHAPTER III. 

THE THREE-HINGED ARCH. 

1. To Pass a Funicular Polygon Through Three Points 126 

2. Determination of the Reactions of a Three-hinged Arch 128 

3. Moments in Three-hinged Arches 131 

4. Influence Lines for Reactions in Three-hinged Arches 133 

5. Influence Line for Stress in any Chord Member of a Three- 

hinged Arch 1 36 

6. Influence Line for Stress in any Web Member of a Three- 

hinged Arch 140 

CHAPTER IV. 

CANTILEVERS. 

1. Definitions 145 

2. Bending Moments in Cantilevers 146 

3. Reaction Influence Lines for CantilEvejis 147 

4. Shear Influence Lines for Cantilevers 150 

5. Moment Influence Lines for Cantilevers 152 

6. Stress Influence Lines for Members of Cantilevers 154 

Cantilevers on Towers 156 

7. Suspended Cantilevers 159 

8. Fixed-load Stresses 159c 

9. Live-load Stresses 1596 

10. Influence Lines for the Chord Members of the Anchor Span 

Li^U 159^^ 

11. Influence Lines for the Chord Members of the Cantilever 

Arm L(,Lf 1S9A; 

12. Influence Line for Web Members of the Anchor Span L^L^ . i^gl 

13. Influence Line for Web Members of the Cantilever Arm L^Lg. 1590 

14. Suspension Cantilever with Two Points of Support at Inter- 

mediate Pier 1593 



X CONTENTS. 

CHAPTER V. 
DEFORMATION OF TRUSSES. 

ART, PAQB 

Introduction i6o 

1. The Displacement or Williot Diagram i6i 

2. Rotation of a Rigid Figure about a Point i66 

3. Deformation of a Bridge-truss 16S 

4. Displacement Diagram for a Three-hinged Aech 173 

5. Displacement Diagram for Truss with Sub-divided Panels.. 175a 



CHAPTER VI. 
THE DETAILED DESIGN OF A RAILROAD BRIDGE. 

Introduction 176 

1. Stresses in the Structure 177 

2. Determination of Dead-load Stresses 178 

3. Determination of Live-load Stresses in the Chords i8a 

4. Determination of Live-load Stresses IN the Web Members ... . 181 

5. Determination of Wind Stresses 183 

Upper Lateral Bracing 183 

Lower Lateral Bracing 184 

6. Design op Lower Chord Members 185 

Member LJL^ 186 

Member LJ^^ 1 87 

7. Design op Upper Chord Members 188 

Member U^U^ 189 

End Post LgU^ 193 

8. Design of Floor-beam Hangers 194 

Member U-^L^ 194 

Members L^M^ and L^^M^ 195 

9. Design of Vertical Main Web Members or Posts 195 

Member UJ^^ ^9^ 

Member U^L^ 197 

Member U^L^ 198 

Tension Member IJ J^^ 199 

ID. Design of Main and Counter Web Members 199 

Member UJ^^ 199 

Member L^^ 200 

Member L^M^ 200 

Member L^M^ 201 

Member MJJ^ 201 



CONTENTS. xi 

AST. PASS 

11. Combined Stresses 202 

Member U^U^ 202 

Member LJL^ 203 

12. Design oe Stringers and Floor-beams 205 

Stringers 206 

Length of Cover Plate 208 

Web Plate 208 

Riveting 209 

Rivets in Cover Plate 211 

Rivets on End Connection 211 

Intermediate Floor-beams 212 

Riveting -•••••. ..•..r«« 314 

13. Specifications for Lateral and Wind Bracing 217 

14. Design of Upper Longitudinal Wind Bracing 218 

15. Design of Portal Bracing 220 

16. Design of Lower Longitudinal Wind Bracing 221 

17. Design of Bedplates, Friction Rollers, and Pedestals 223 

18. Design of End Floor-beam 226 

19. Design of Pins and Joint Details 227 

Pin-plates of Posts 231 

Member LJJ.^ — Lower End 231 

Meynber LJJ^ — Upper End 232 

Member UJ^^ — Lower End 232 

Member U^L^ — Upper End 233 

Member U^L^ — Lower End 233 

Member U^L^ — Upper End 233 

20. Design of Pin Connections of Upper Chord Members 234 

Connection at U^ 235 

Connection at U^ 237 

21. Design of Details at Ends of End Post 237 

Lower Extremity 237 

Upper Extremity 238 

End Connection of UJJ^ 238 

22. Bending of Pins 239 

Lower Chord Packing, Panel Point Lj 239 

Upper Chord Packing, Panel Point U^ 241 

23. Camber 244 

24. Conclusion 245 



GRAPHIC METHODS FOR BRIDGE AND 
ROOF COMPUTATIONS. 



CHAPTER I." 

GENERAL PRINCIPLES OF GRAPHIC STATICS, AND APPLI- 
CATION TO ROOF-TRUSSES. 

Graphic statics is an adaptation of graphical methods 
to the study of statics, so that the values of quantities 
otherwise found algebraically may be determined by the 
aid of geometrical diagrams or figures. 

Art. I. — Definitions. 

A force is completely known when its magnitude, 
direction, and point of application are given, but the line 
of action without the point of application is generally 
sufficient. So far as any problems in statics are concerned, 
a force may be considered as acting at any point along 
its line of action. The location of the point of application 
of a force affects only the material upon which the force 
acts. 

The magnitude, direction, and line of action of a force 
may be represented by a straight line, the length of the 
line representing the magnitude by a proper scale. The 
pointing of the line with an arrow-head, or a proper reading 
of the letters placed at the ends of the line, will indicate 



2 GRAPHIC STATICS. Ch. I. 

the direction of the force. In Fig. i the force P acting 
in the direction of the arrow must be read ah and not ha. 

Forces may be coplanar or non-coplanar, and either 
concurrent or non-concurrent. Coplanar forces have Hnes 
of action all in one plane, while non-coplanar forces may 
have lines of action anywhere in space; for the present 
the latter will not be considered. Concurrent forces have 
lines of action all meeting in a point, but the lines of action 
of non-concurrent forces do not meet in a single point. 

The resultant of any system of forces is a single force 





which, having the same effect as to equilibrium or motion 
as the system itself, may replace it. 

In accordance with the principles of composition and 
resolution of forces, if Pi and P2 (Fig. 2) are two concur- 
rent forces acting at A in the directions BA and CA re- 
spectively, and if P2 be transposed so that CA becomes 
the side DB of the triangle ABD, then will DA represent 
the resultant in magnitude and direction with its point of 
application at A. 

It is seen that in following around the sides of the 
triangle, the resultant must be taken in a direction opposite 
to that of its component forces. The forces AB and AC 
are said to be the components of the force AD, and by 
drawing other triangles upon ^Z) as a base an indefinite 
number of pairs of other components may be found. Com- 
ponents are called rectangular when they are at right 
angles with each other. It should be carefully noted that 
the direction of a force is always given by the proper 



Art. 2.] RESULTANT OF ANY NUMBER OF FORCES. 3 

reading of the letters placed at the ends of the line repre- 
senting it. 

Art. 2. — Resultant of any Number of Forces. 

The method of finding the resultant of two concurrent 
forces is easily adapted to finding the resultant of any 
number of concurrent forces. Let the forces P\, P2, 
Pz . . . Pi (Fig. 3) act at A, and let them be transferred 
to Fig. 4, so as to form the sides of the polygon BCD. . . , 
the forces being directed around the polygon in the same 
way. It is seen that Ri is the resultant of Pi and P2, 
its direction in the triangle BCD being from B to D. Simi- 







Fig. 3. Fig. 4. 

larly R2 is the resultant of Ri and P3, and finally R or BF 
is the resultant of all the given forces, and it acts in a direc- 
tion opposite to that of the other sides of the polygon. 

In order to find the resultant of any number of con- 
current forces, it is necessary only to lay down the forces 
as the sides of an open polygon, their directions following 
around the polygon in the same sense. The closing side 
of the polygon will then represent the resultant in magni- 
tude and its direction must be opposite to that of the 
other forces. The polygon may have re-entrant angles, 
or the sides may cross, but the form of the figure does 
not affect the result. The order in which the forces are 
laid down is also a matter of indifference, but to avoid 
confusion they are usually taken consecutively about any 
point. 



4 GRAPHIC STATICS ■ [Ch. I. 

A force may be held in equilibrium by another force, 
equal to it in magnitude and having the same line of action, 
but opposite in direction. Therefore if the direction of 
the resultant found in Fig. 4 be reversed, the system of 
forces will be held in equilibrium by FB, since FB exactly 
balances BF. Any system of concurrent forces, therefore, 
is in equilibrium if the separate forces may be laid down 
as the sides of a closed polygon and if the lines of action 
of these forces follow around the polygon in the same way. 
Such a figure is called a force or equilibrium polygon, and 
it represents graphically two of the three equations of 
condition for equilibrium of coplanar forces, viz., first, that 
the sum of the horizontal components of the forces must 
equal zero, and secondly, that the sum of the vertical 
components must equal zero. 

Art. 3. — Application of Force Polygon to Finding of Unknown 

Quantities. 

Since the force polygon represents two equations of con- 
dition, the finding of two unknown quantities in a system of 
concurrent forces in equilibrium is made possible. Therefore 
if all the forces in such a system are known except two, 
and if the lines of action of these two be given, the un- 
known quantities may be found by means of the force 
polygon. Let P4 and P5 (Fig. 5) represent the lines of 
action of these two unknown forces, the forces Pi, P2, and 
P3 being fully known. If the force polygon (Fig. 6) for 
Pi, P2, and P3 be drawn and if the figure be closed by 
drawing lines parallel to P4 and P5 at the open ends of the 
polygon, P4 and P5 will be completely determined. It 
is obvious that two solutions leading to the same results 
are possible, one of which is shown by the broken lines. 
In using the force polygon it will be found convenient to 
take the forces in the order of their rotation as they act 



Art. 4 J SYSTEM OF NOTATION. 5 

about the point of application, although that is evidently 
not necessary. 

In a similar manner the force polygon furnishes a 
solution in the case of a system of concurrent forces, if 
all the forces except one are completely known. In that 
case the closing side of the polygon fully determines the 
resultant in amount and direction. 

It is evident that the two equations of condition for 
equilibrium represented by the method of the force polygon 





Fig. 5. Fig. 6. 

must also be applicable to non-concurrent forces; but a 
third equation of condition must also hold, viz., the sum 
of the moments of all the forces taken about any point in 
their plane must be equal to zero. The graphical treat- 
ment of this condition will be considered later. 

Art. 4. — System of Notation. 

The ease of operating with graphical methods is due 
in a great measure to the system of notation invented 
by Bow, and kriown by his name. Two diagrams are 
employed, one showing only the lines of action of the 
forces in connection with the structure, and called the 
truss diagram, while the other is simply the force polygon 
already explained. 

The space between the lines of action of any two adja- 
cent forces on the truss diagram, or the space on that dia- 
gram within any polygon whose sides are formed by the 



6 GRAPHIC STATICS. [Ch. I. 

lines of action of any forces, is marked by a capital letter. 
Each force will then always lie between two and only 
two such capital letters, and consequently it can be desig- 
nated conveniently by the two letters between which it 
lies. These letters, changed from capital to small letters, 
are then placed on the force polygon at the ends of the 
lines representing the same forces. The correct disposition 
of the letters at the ends of the lines indicates the direction 
of action of each force, as has already been explained. 

Example. 

Fig. 7, representing a simple king-post truss carrying a 
concentrated load at its centre, illustrates the truss dia- 
gram. The external forces acting upon the truss are the 
two reactions at its ends and the concentrated centre 
loading. The space between the concentrated load and 
the left reaction is marked by the letter A, the space above 
the truss and between the reactions by B, and the space 
between the concentration and the right reaction by C. 
The apices of the triangles are numbered i, 2, 3, and 4. 
According to this system of notation the left reaction will 
be known as ah or ba, the load as ac or ca, and the right 
reaction as ch or he. The indetermination due to the use 
of two names for one force is made to vanish by the use 
of the following rule : In any truss diagram all forces must 
be read in the same order, clockwise or counter-clockwise, 
around the figure or about a point. 

In Fig. 7, employing the direction indicated by the 
circular arrow placed in the upper left-hand corner, the 
names of the three external forces are then immediately 
fixed, as ah, he, and ea. 

The internal forces, or stresses in the members of the 
truss, are designated in exactly the same manner, the 
triangular spaces in the truss being marked D and E. The 
forces acting about the point i are then hd, da, and ah; 



Art. s.] 



DETERMINATION OF KIND OF STRESSES. 



those about point 2, ed, db, and be; etc. It is seen that, 
the same force has a different name when read with respect : 
to two different points; thus the member between panel ■. 
points I and 2 is the member or force bd with respect to 
point I, and db with respect to point 2. As will be shown 
presently, this feature of the method determines at once, 
the character of the stresses in the various members. 



Art. 5. — Det3rmination of Kind of Stresses. 

Since the structure is in equilibrium under the loading 
assumed, every panel point is also in equilibrium, and 
consequently all the forces at any one such point are in 
equilibrium. A force polygon may therefore be drawn 
for the concurrent forces acting at each panel point. 

The reactions in Fig. 7 may each be found by analytical 
methods to be equal to one half the centre concentration. 





Fig. 7. 



Fig. 8. 



There remain, then, at point i only two concurrent forces 
with known lines of action to be determined. The force 
ab, since it acts upwards, is laid down on the force diagram 
(Fig. 8), with the letter a at the bottom of the line repre-. 
senting its magnitude and direction. The line bd is then 
drawn parallel to BD, beginning at b, and the line da 
parallel to DA , beginning at a ; their intersection will fix . 
the point d. Since for equilibrium the forces in this force 
polygon must follow each other in the same sense, and; 



S GRAPHIC STATICS. [Ch. I. 

since ab is fixed, the direction of the force hd is from b 
to d; transferring this direction to the truss diagram, it 
will be seen that the force bd acts towards the point i . 

It is clear that a force acting towards a point, in the 
manner of bd, represents compressive stress in the member 
in which it acts, in the same manner that a force acting 
away from a point represents tensile stress. In the present 
case bd therefore acts in compression with a magnitude 
represented by the length of the line bd. Similarly, trans- 
ferring the direction of da to the truss diagram, it will be 
found that da acts away from i and represents a tensile 
stress. Thus all the forces acting at the point i have been 
fully determined. 

Other panel points at which there are but two unknown 
forces must next be considered in succession. 

At panel point 2 the forces must be read db, be, and ed, 
bd being already drawn on the force diagram. A line 
parallel to BE is therefore drawn through b and one parallel 
to ED through d ; then the intersection of these lines furnishes 
e on the force diagram. By transferring the direction of 
be to Fig. 7, it will be found acting towards panel point 2, 
causing compression. Similarly ed acts downwards from 
point 2 and causes tension. 

Panel points 3 and 4 are to be treated similarly, but 
it is unnecessary to explain the operations in detail. 

Art. 6. — Crane-trusses. 

The graphic principles thus far explained may be 
illustrated by an application to a form of truss which has 
been used for powerful cranes under circumstances re- 
quiring much headroom, and shown in Fig. 9. The truss 
revolves about a vertical axis situated in the centre of 
the member BD. In that example the weight hanging 
ftom the peak is supposed to be 20,000 pounds. Each 



Art. 6.] 



CRANE-TRUSSES. 



chord of the truss NA, LA, J A, etc., or MC, KC, IC, etc., 
is made up of chords of quadrants of two circumferences 
of circles, the radius for the inner chord being 23.5 feet, 
and that for the outer chord 26 feet; the member BD 
is 5 feet. 

The lengths of the various members are as follows : 

NA = 5 feet ; 
LA=JA=MC= 6 feet; 
HA^KC^-] ieet; 
FA=IC = 8 feet; 
GC = g feet. 



,4^3.1-^ 




Fig. g 



Fig. 10 is a complete diagram for the stresses in the 
truss, supposing the only load to be the 20,000 pounds 
hanging from the peak. If it should be necessary to take 



lO GRAPHIC STATICS. [Ch. I. 

into account the weight of the truss itself, it would be done 
by considering the weights of the various members as 
concentrated at the different panel points, and the problem 
would be treated precisely as in the case of roof -trusses, to 
be treated in succeeding articles. 

The construction of Fig. lo is begim by considering the 
equilibrium of panel point i ; the line ca is drawn downward 
parallel to the line of action of the weight CA, and then at 
its ends are drawn lines parallel to AN and NC The order 
in which forces are taken about any panel point is indicated 
by the direction of the arrow shown in the upper left-hand 
comer of Fig. 9. Panel point i is the first panel point at 
which the stresses can be determined, since at that point 
there are but two unknown quantities. Other panel points 
are then to be taken in such order that at the point con- 
sidered there are only two unknown forces. This is imper- 
ative. The stresses at panel point 2 are therefore treated, 
the lines nm and mc being drawn respectively parallel to the 
corresponding members in the truss diagram. The other 
panel points m.ay then be treated in exactly the same 
manner, considering them in the order in which they are 
numbered. 

In order to determine the character of the stress in 
any member, such as AIN, it is only necessary to observe 
that the direction of mn in the force diagram is downward 
and toward panel point 3. It is in connection with panel 
point 3 only that the micmber MN is named in that way, 
and the latter is therefore in compression. If this member 
were to be named in regard to panel point 2, it would be 
termed NM. Its direction would then be upward with 
regard to panel point 2, indicating compression as already 
found. 

The diagram gives the following results, the positive 
sign indicating tensile stress, and the negative sign com- 
pressive: 



Art. 6.] 



CRy^NE-TRUSSES. 



II 



Stresses in the Crane of Figure 9. 


Upper Chord. 


Lower Chord. 


Web Members. 


CiV= +26,000 lbs. 


AN = — iQ,ooo lbs. 


NM= —27,000 lbs. 


CM= +34,000 lbs. 


AL= —50,000 lbs. 


ML= + 15,000 lbs. 


Ci?:= +62,000 lbs. 


AJ = — 73,000 lbs. 


LK= —29,000 lbs. 


C/= + 79,000 lbs. 


AH= —90,000 lbs. 


KJ = + 3,000 lbs. 


CG= +91,000 lbs. 


^i^=— 99,000 lbs. 


// = —27,000 lbs. 


CE= + 102,000 lbs. 


AD= — 112,000 lbs. 


IH=— 4,000 lbs. 
HG= —28,000 lbs. 


Reactions. 




GF= — 15,000 lbs. 


BC = I 94,000 lbs. 




FE = — 34,000 lbs. 
ED= — 12,000 lbs. 


AB= 1 104,000 lbs. 




BD = — 9,000 lbs. 



These results can easily be checked by the algebraic 
method of moments. For instance, the stress in the member 
IH is found by taking a section through the truss as shown 
(Fig. 9) and taking moments about the intersection of CI 
and HA ; the lever-arm of IH as scaled from the diagram is 
15.4 feet. The only external force to the right of the section 
is the weight at the peak, which, by scale, has a lever-arm 
about the centre of moments of 3.1 feet. The equation of 
moments is then as follows, the direction of IH being indi- 
cated by the arrow lying along it: 

— ///X 15.4 = 20,000 X 3.1 ; 
.'. ///=— 4000 pounds. 

This value checks that obtained by the graphical method. 
The last force found graphically is the reaction BC, and 
if its graphical value be checked by the method of moments, 
it may be assumed that all previous graphical results are 
correct. For this member the centre of moments is at 
panel point 1 3 ; the equation of moments is 

BCx^ =20,000X23.5; 
.'. £C* = 94,000 pounds. 

This resiilt agrees with the value obtained from the dia- 
gram. 



12 



GRAPHIC STATICS. 



[Ch. L 



If a chain, rope, or cable pass along either chord, the 
tension in it will tend to produce an equal amount of com- 
pression in the panels of that chord; the resultant stress, 
therefore, in any panel will be the algebraic sum of this 
amoiant of compression and the stress due to the weight at 
the peak. 

Fig. II is a skeleton diagram of an ordinary crane, 
which revolves about the centre line of 2-3 as an axis. 
AB is the weight hung at the peak, i . The force diagram. 
Fig. 12, is found by drawing ah vertically downwards equal 
to the weight, then at a and h drawing he and ca respectively 
parallel to BC and CA of Fig. 11. he then represents the 





Fig. II. 



Fig. 12. 



compression in BC, and ea the tension in CA. As before, 
the tension in the rope or cable tends to produce an equal 
amount of compression in any member along which it 
lies. 

Let / denote the normal distance from the line of action 
of CA to any point in the centre line of the vertical member 
2-3 ; then any section of 2-3 will be subjected to the bend- 
ing moment 

M=ca-l, 



ca representing the stress in CA. 



Art. 7.] 



NON-CONCURRENT COP LAN A R FORCES. 



13 



2-3 will also be subjected to a direct stress (tension in 
Fig. 11) equal to the vertical component of the stress in 
CA. 

The greatest resultant intensity of stress in any section 
will be the combination of the intensities due to these two 
causes. 

Art. 7. — Non-concurrent Coplanar Forces, and the Funicular 

Polygon. 

Concurrent forces only have thus far been treated, but 
the principles employed in the concurrent force diagram 
are equally applicable to non -concurrent coplanar forces. 




Fig. 13. Fig. 14. 

In the case of equilibrium of such forces, one other condi- 
tion must also hold, viz., the sum of the moments of these 
forces taken about any point as a centre must be equal to 
zero. This condition may be expressed graphically, as 
will presently be shown. 

Let Pi, P2, Pz, and P4 (Fig. 13) represent four non- 



14 GRAPHIC STATICS. [Ch. I. 

concurrent forces whose resultant is to be completely deter- 
mined. By combining Pi and P2 in the force diagram, 
Fig. 14, their resultant is found to be -Ri and its line of 
action passes through a at the intersection of Pi and P2 
in Fig. 13. This resultant may then be combined with 
P3 (Fig. 14), producing i?2, whose line of application passes 
through the point h (Fig. 14), the intersection of Ri and P3. 
Similarly, this resultant may be combined with P4, pro- 
ducing the final resultant R (Fig. 14), whose line of action 
passes through the point c (Fig. 13). 

This method becomes inapplicable when the forces 
treated are parallel or nearly parallel, but the following 
construction, which is perfectly general, may then be 
employed : 

Let Fig. 15 represent five forces Pi, P2 . . . P5, whose 
resultant it is desired to obtain; Fig. 16 shows the value 
and the direction of its resultant as obtained by the usual 
force polygon, but both its point of application and line 
of action (Fig. 15) are still unknown. Let be any arbitrary 
point in Fig. 16, and let there be drawn from to the ends 
of all the forces Pi, P2 . . . P5 radiating lines numbered i, 2, 
3 ... 6. It is then evident that in Fig. 16 each force has 
been resolved into two components, as Pi resolved into 
I and 2, and P2 into 2 and 3, and so on. Choosing any 
point on Pi, such as a in Fig. 15, let the line 2 be drawn 
parallel to the line 2 in Fig. 16 until it intersects the line 
of action of the force P2 ; from this intersection point draw 
3 parallel to 3 in Fig. 16 until it intersects P3, and so con- 
tinue the construction to its completion. Finally produce 
the lines i and 6, Fig. 15, to their intersection at the point 
h. This is a point in the line of action of the resultant; 
for at point a in Fig. 15 the force Pi has been replaced 
by the force components i and 2 acting in the directions 
indicated by the arrows, these directions being shown in 
Fig. 16. P2 has similarly been replaced by the forces 2 



Art. 7.] 



NON-CONCURRENT COPLANAR hORCES. 



15 



and 3 acting in the directic;ns similarly shown; but it is 
evident that the two forces 2 between Pi and P2 balance 




Fig. 15. 




Fig. 16. 



each other since they are equal in amount, opposite in 
direction, and act along the same line. Similarly, the two 
forces 3 between P2 and Pz cancel, as do likewise the pairs 



1 6 GRAPHIC STATICS. [Ch. I. 

of forces 4 and 5. The five forces Pi, P2 . . . Pa have then 
finally been replaced by the two forces i and 6. These, 
as is evident from Fig. 16, are the components of the final 
resultant, and may therefore be replaced by the resultant, 
which must act at their intersection b. 

The polygon 1-2-3-4-5-6 drawn in Fig. 15 is termed 
the polygonal frame or funicular polygon, because it is a 
frame of members or bars, which, if of sufficient sectional 
area, will sustain the set of external forces Pi, P2 ■ ■ ■ P& 
as long as their lines of action and magnitudes remain 
unchanged. The term funicular polygon is so distinctive 
that it will hereafter be invariably used for this polygon, 
although the term polygonal frame is much used. The 
point from which the radial lines are drawn is termed 
the pole, and it is usually designated by the letter 0. The 
distance from to i? is termed the pole distance and is 
generally represented by H. The radial lines, which are 
always designated by numbers, are called rays. 

. These closed diagrams fulfil the condition that the 
sum of the moments of the forces of the system about any 
point must equal zero, as will now be shown. 

Let it be determined to find the sum of the moments 
of all the forces Pi, P2 ■ ■ ■ P5 about the point N, which 
is distant x from their resultant R in Fig. 15. Since the 
moment of the resultant about this point is equal to the 
sum of the moments of its various components, it will 
be sufficient to treat only that resultant. The moment 
is, therefore, M=R.x. Draw through the centre of 
moments A'' the line cd parallel to the resultant, and con- 
tinue the lines i and 6 until they intercept on cd the dis- 
tance /. The triangles bed and OAB (Fig. 16) are similar 
since their sides are respectively parallel. The following 
proportion is therefore obtained: 

x:cd::H:R, 



Art 8.] SPECIAL FEATURES OF THE FUNICULAR POLYGON. 17 

or 

M=R.x = cd.H. 

Hence the moment of all the forces about the assumed 
centre, N, is equal to the product of the intercept on the 
line (Fig. 15) drawn through that centre parallel to the 
resultant between those rays which are the components 
of R, multiplied by the pole distance. More simply stated, 
the moment is the product of an intercept on the truss 
diagram multiplied by the pole distance of the force 
diagram. 

It is seen that this moment reduces to zero when the 
centre of moments is taken at the intersection of the 
two component rays of R\ viz., at the point h of Fig 15, 
or if taken anywhere on the line of action of R, for in 
either case the intercept equals zero. 

If the five forces P\, P2 . . . P5 form a system in equilib- 
rium, the force polygon in Fig. 16 will close and the re- 
sultant R will be zero, making the moment AI =R .x = o. 
Hence the sum of the moments of all the forces of the 
system about any point N will be zero. Again, in Fig. 15, 
if the direction of R be reversed, it and the five forces Pi, 
P2 . . . P5 will constitute a system in equilibrium. In 
that case the funicular polygon becomes a-b-c- ^-4.-^2 and 
it is closed. If a system of forces is in equilibrium, there- 
fore, both the force polygon and the funicular polygon 
or frame will be closed. These two polygons therefore 
express graphically the three static equations of condition 
for equilibrium of non -concurrent forces. 

Art. 8. — Some Special Features of the Funicular Polygon. 

Although the funicular frame in Fig. 15 was drawn by 
beginning at a in the line of action of Pi, it is now evident 
that any point on the line of action of any force might 
have been taken as a point of beginning, while the pole 



1 8 GRAPHIC STATICS. [Ch. L 

and the rays remain unchanged, and that for each such 
point a different funicular polygon would have been found. 
With the same pole and the same set of rays, therefore, 
there may be an infinite number of funicular polygons for 
the same system of forces in equilibrium; but since the 
rays are unchanged the corresponding sides of all those 
polygons will be parallel, or, in other words, those polygons 
or frames will be concentric. 

Again, any point whatever may be taken as the pole 
in Fig. 1 6, and rays may be drawn from any such point 
to the angles of the force polygon. Hence there are an 
infinite number of poles for each of which there an an 
infinite number of concentric funicular polygons for any 
given system of coplanar forces ; but when the position of 
the pole is changed the corresponding sides of the polygons 
or frames generally cease to be parallel to each other. The 
pole, therefore, may be either within or without the force 
polygon, or at any point of its perimeter. 

If the pole is located at an angle of the force polygon, 
as at B, Fig. i8, the ray drawn to that angle will become 
zero in length and the corresponding side of the funicular 
polygon, as 5 in Fig. 17, will disappear. In other words, 
there will be no stress in the side 5 of Fig. 17, and the 
directions of P5 and P4 will coincide with the directions 
of sides I and 4 respectively, so that the side or bar 5 would 
be omitted. This case is that of the ordinary masonry 
arch. 

When the pole is changed in position on the force dia- 
gram, the pole distance is altered. Since the pole distance 
appears directly as a factor in determining the moments 
of any of the forces in the diagram, it follows that with 
an increase of pole distance there will be a decrease of 
intercept on the funicular frame; that is, the intercept 
varies inversely with the pole distance. This consideration 
is of importance in the treatment of arched ribs, for in those 



Art. 8.] SPECIAL FEATURES OF THE FUNICULAR POLYGON. 19 

structures the pole distance represents the horizontal 
thrust. 

If the pole is located in some side of the force polygon, 
as at some point in P5 of Fig. 18, two rays will be coin- 
cident in direction with that side, and the two sides or 
bars of the funicular polygon, as i and 5 in the case sup- 
posed, will coincide in direction with each other and with 
the force P5 acting at their common extremity. The funicu- 
lar polygon will thus apparently have one less side or bar 
than the number of rays in the force polygon. 




Fig. 17. 



Fig. 18. 



Remembering the double significance of the rays as 
representing the stresses in the bars of the funicular poly- 
gon by the same scale as that to which the forces J^i, P2, 
etc., are measured, it is well to observe that each side of the 
force polygon with the two rays adjacent to it form a tri- 
angle of forces in equilibritim, which three forces are those 
meeting at the corresponding angle of the funicular polygon. 

In Fig. 17 the external forces are taken as acting inward 
or towards the frame; but all directions may be reversed 
and the funicular polygon may be inverted without changing 
in any respect any of the preceding demonstrations, con- 
structions, or conclusions. If the force directions are 
reversed and the funicular polygon or frame inverted, the 
condition shown in Fig. 19 will result, which, if P5 and P4 



20 GRAPHIC STATICS. [Ch. I. 

coincide in direction with i and 4, so that 5 disappears, 
is that of the elementary suspension-bridge cable. 

It is to be borne in mind that thus far the system of 
forces supported in equilibrium by the funicular polygon 
has been supposed to be invariable in every respect. The 
extreme forces P5 and P4 are usually reactions or support- 
ing forces supplied by piers or abutments. 




Since the force and funicular polygons correspond to 
three equations of condition of statics, they may therefore 
be used for finding not more than three unknown quanti- 
ties, in any problem involving the equilibrium of coplanar 
forces. Their application to the general solution of some 
problems of statics may therefore be discussed. 

Art. 9. — Lines of Action of all Forces Known, but Magnitudes 
and Directions of Two Forces Unknown. 

In Fig. 20 let the lines of action of all the forces Pi . . . P5 
be knov/n, but let the magnitudes and directions 8i any 
two, as P4 and P5, be unknown. Since it is a matter of 
indifference in what order the forces are taken in the con- 
struction of the force polygon, let the known forces be taken 
in order from Pi to P3, as shown. Then the known lines 
of action of P4 and P5 will enable the two remaining sides 



Art. 9 ] SOLUTIONS IVITH AID OF THE FUNICULAR POLYGON. 21 



of the force polygon to be drawn from A and B, while the 
directions of those two forces will be determined by travers- 
ing the perimeter of the polygon in the way or sense indi- 
cated by the known sides, all as shown in Fig. 21. The 
funicular polygon can then be drawn from any point in 
any line of action in the manner already established in 
the previous articles. 

Although the unknown forces are shown in Fig. 20 as 
adjacent to each other, it is clear from the demonstra- 
tion that they may be any two whatever of the system. 

If the unknown forces are parallel to each other, their 
intersection (that of P4 and P5 in the force polygon of 





Fig. 20. 



Fig. 21. 



Fig. 21) in the force polygon becomes indeterminate. In 
that case it is only necessary to complete the funicular poly- 
gon before completing the force polygon, as is illustrated 
by Fig. 22, in which all the forces are supposed to be 
parallel to each other. The forces whose lines of action are 
known but whose magnitudes and directions are unknown 
.are P4 and P5. The force polygon now becomes a straight 
line, since all the forces are parallel, and in order to show 
completely this coincidence of force lines all the forces 
in Fig. 23 are represented by close parallel lines. The 
force (full) lines Pi, P2, and P- are first drawn m any 
order that will give their proper consecutive directions, 



22 



GR/iPHlC STATICS. 



[Ch. I. 



and the rays i, 2, 3, and 4 are then run from any pole O 
to the extremities of the former. By commencing at any 
point in any of the Hnes of action (all known) the four rays 
will enable all sides or bars of the funicular polygon, except 
5, to be drawn in the usual way. Both extremities A and 
B of the side 5 will thus be fixed and the line itself can at 
once be completed. The ray 5 can then be drawn from 
the pole to the point C in the straight line which repre- 
sents the force polygon. The forces P4 and P5 will be 
represented in magnitude by lines extending from C to 
the extremities of P3 and Pi, and their directions (P4 down 





Fig. 22. 



Fig. 23. 



and P5 up) will be found by traversing these lines in the 
same sense as the known force line Pi . . . P3. Both forces 
and frame are thus completely determined. 

It is of the greatest importance to observe that the force 
polygon for a system of parallel forces reduces to a straight 
line, which is really composed wholly or in part of two 
or more coincident straight lines. It is also to be noted 
that any pole has been chosen, but that any other of 
the infinite number of possible poles would lead to pre- 
cisely the same final results. This latter case of the parallel 
-forces is a common one in practice, but usually the two 
extreme forces are unknown in magnitude only. 



Art. ic] MOMENTS IN A BEAM. 23 

It may be observed that if the directions of the two 
component rays of either unknown force in the force polygon 
of Fig. 23 and the magnitude of one of those components 
are known, the intersection of the line of action of the 
other component with the force line CD will fix the point 
C and determine the unknown forces without the aid of the 
funicular polygon. 

Art. 10. — Moments in a Beam. 

The funicular polygon in the case of a beam carrying a 
system of parallel loads represents the bending moment 
at every section of the beam. Let Fig. 24 represent a 
simple non-continuous beam carrying the loads Pi, P2 • • ■ -P5 
at the points shown. This system of weights is held in 
equilibrium by the two reactions R and R\, whose directions 
and points of application are known, but whose amounts 
are to be determined by the funicular polygon. 

Lay off in regular order the forces Pi, P2 . . . P5 in the 
force diagram Fig. 25, and choose any point as a pole. 
Draw the rays i, 2 ... 6 and transfer them in the proper 
manner to Fig. 24. This funicular polygon can only be 
drawn for the bars i, 2, 3, 4, 5, and 6, for this disposes 
of all the. known rays ; but since the polygon must close, 
the side 7 is drawn as the closing line and transferred to 
Fig. 25. Its intersection with the line of loads at m deter- 
mines the amounts of the reactions R and Pi as shown. 

The moment at any section of the beam, such as xy, 
may then be obtained by the rule of the ' ' product of inter- 
cept b}'' pole distance" (Art. 7). The resultant of the 
forces P, Pi, P2, and P3 to the left of the section is included 
between the two rays 4 and 7 of Fig. 25 ; consequently the 
intercept required lies between the bars 4 and 7 and is the 
line ah in Fig. 24. The moment at xy is therefore ah . H, 
where H represents the pole distance (not shown) . For par- 
allel forces the pole distance is a constant quantity, and is 



24 



GRAPHIC STATICS. 



[Ch. I. 



generally chosen as a unit force, frequently looo or 10,000 
pounds, in order to avoid arithmetical computation. In 
general, then, the moment at the section xy is expressed by 
the ordinate directly below it, if the significance of the pole 




Scale in F^ct 



Fig. 24. 

distance be remembered. The moment at any other 
section, such as x'y', is then also expressed by the ordinate 
below it, or a'b' . 




Scale in Pounds 



Fig. 25. 

The equivalence of the funicular polygon as a moment 
polygon applies as well to cantilever or continuous beams 
as to non-continuous beams, provided that the loading 
consists of a series of parallel weights. 



Art. II.] SOLUTIONS IVITH AID OF THE FUNICULAR POLYGON. 25 



Art. II. — All Forces Given Except Two, of which the Line of 
Action of One and the Point of Application of the Other 
are Known. 

This case is represented by Fig. 26, in which Pi ... P^ 
constitute any system of forces all except two of which, 
as Pi and P5, are known. The line of action of one un- 
known force, F5, is given together with the point of appli- 
cation A of the other. That portion of the force polygon 
representing the known forces Pi ... P^ is first drawn in 
the usual manner, together with the rays i to 4, inclusive, 
from any pole 0, Fig. 27. Then, by starting at the known 




Fig. 26. 



Fig. 27. 



point of application A, the sides or bars of the fimicular 
polygon or frame can be drawn parallel to the corresponding 
rays, so as to fix the point B on the line of action of P^. 
The last side 5 of the frame (Fig. 26) will thus be completely 
determined ; and if the ray 5 (Fig. 27) then be drawn parallel 
to it until it intersects the line drawn through C parallel 
to the line of action of Pg, the latter force and P4 will at 
once become completely known. 

This case is one of common occurrence, the forces P4 
and P5 being in general the reactions or supporting forces 



26 



GRAPHIC STATICS. 



[Ch. I. 



of the structure at the abutments or piers. It is obvious, 
however, that any other two forces besides the reactions 
or extreme forces might have been taken as the unknown 
elements. 

If the forces are all parallel, the force polygon Pi ... P^ 
becomes a straight line, as already explained in Fig. 23, 
but the process of solution and the various steps in the 
construction remain unchanged in every particular. 



Art. 12. — All Forces Given Except Three, with the Lines of 
Action of those Three Known, 

Let the six forces Pi . . . Pq in Fig. 28 represent any 
system of forces whatever, of which all lines of action are 
known, but the magnitudes and directions of P2, Pz, 




/V---? 



/ ^ \ 

/5 \Po \P 




Fig. 28 



Fig. 29. 



and Pq remain to be determined. The points of applica- 
tion of the forces Pi, P2, Pz, Pi, P5, and Pq are supposed 
to be a, b, c, d, e, and / respectively, but their lines of action 
will be prolonged and used for the purposes of this solution. 
So much of the force polygon as can be constructed from 
the known forces Pi, P5, and P4 is first drawn, and then 



Art. 12.] SOLUTIONS IVITH AID OF THE FUNICULAR POLYGON. 27 

the four rays i, 2, 3, and 4 are drawn from any pole O. 
The point ^4 (Fig. 28) of intersection of any two Hnes of 
action (as those of P2 and P3) of the unknown forces is 
next taken as the starting-point of the funicular polygon. 
The rays drawn to the angles of the partially constructed 
force polygon will at once enable the sides 4, 3, 2, and i 
of this polygon to be drawn as shown. The points A and 
B of the remaining side -5' are thus fixed in Fig. 28, and the 
side itself completed. The ray 5 is then drawn (Fig. 29) 
until it intersects the line of Pe drawn from E. The points 
C and D are thus the starting-points from which P2 and 
P3 are to be drawn until they intersect and so complete 
the force polygon Fi, Pq, P2, Ps, Pa, Pb- The latter give 
the magnitudes and directions of all three of the unknown 
forces. 

By locating an angle of the frame at a point of inter- 
section of the lines of action of two of the unknown forces 
the sixth side of the frame is reduced to zero and eliminated 
from consideration, so that the corresponding ray is not 
needed. Obviously this method cannot be used with 
parallel forces, for the reason that there will be no finite 
point of intersection A. 

In case the point of intersection A is found to be in- 
accessible or outside the limits of a convenient diagram, 
the usual procedure is shown in Fig. 30. CD and AB are 
the lines of action of P2 and P3 respectively, and their 
intersection is inaccessible. Any point, as E on the adja- 
cent line of action P4, is taken as the starting-point of the 
funicular polygon, the pole not yet having been determined. 
The side 4 of the frame is to be drawn from E to the in- 
accessible point of intersection. Any points A and C 
on AB and CD are so taken as to form the triangle ACE 
with good intersections. From any other point D on CD 
and at an appropriate distance from C draw DB parallel 
to CA and then from D and B draw lines parallel to CE 



28 



GRAPHIC STATICS. 



[Ch. I. 



and Ah respectively, locating the point F by their inter- 
section. EF produced will then pass through the inter- 
section of AB and CD. EF will then be a portion of the 
side 4 of the polygonal frame, and the ray 4 must be drawn 
parallel to it from D in the force polygon of Fig. 29; the 
pole may then be located anywhere on that ray. The 
remaining rays are drawn to the known angles of the 
force polygon and the construction is made as before. 
When the point B, Fig. 28, is reached, the remaining side 
5 of the frame is to be drawn to the inaccessible point of 




^A 


^ /^ 




/ 

/ 
/ 
/ 
/ 




/ 




\ / 


\ 




V 


\ 
V 



Fig. 30. 



Fig. 31. 



intersection of P2 and P3 by the process given in connection 
with Fig. 30. 

If there are but four non-parallel forces in the system,* 
a frequent procedure is that shown in Fig. 31. P2 is the 
only completely known force, the lines of action, only, of 
the others being given. The four forces may be divided 
into pairs, as Pi and P2, and F3 and P4, the resultant of 
each of which pairs must, for equilibrium, be equal and 
opposite to the resultant of the other. Hence construct 

* All systems of the kind treated in this article may be reduced to four non- 
parallel forces; for all the known forces may be replaced by their resultant, and 
there are then in addition to this resultant only the three unknown forces. It 
may be necessary, however, to draw a funicular polygon to determine the point 
of application of this resultant. 



Art. 13. J STRESSES IN ROOF-TRUSSES. 29 

the force triangle Pi, P2, and the resultant ab. As Po 
is completely given, this can be done and both Pi and 
the resultant ab will be determined. Then lay off cd=ab 
from c and construct the force triangle with P3 and P4, 
which will completely determine those forces and make 
known all elements of the problem. 

The operations of this and the preceding articles are 
designed to illustrate the general character of the pro- 
cedures required for the solution of problems by the aid 
of the force and funicular polygons; they include some 
of the more important practical cases, although others 
may arise which are to be treated in the same general 
manner. 

Art. 13. — Stresses in Roof -trusses. 

Prelmiinary. 

Roof-trusses are designed to carry, first, their own 
dead weight ; second, the weight of the roof covering on 
the trusses ; third, a snow load ; fourth, a wind load acting 
in a horizontal direction, first from the right, and then from 
the left; and, fifth, a ceiling or other suspended weights, 
such as cranes, trackways, shafting, etc. 

The weight of the roof-truss itself varies, naturally, 
with the span length and with the distance between trusses. 

Various formulas have been proposed for determining 
the weights of roof-trusses, and among them is the follow- 
ing, deduced by Milo S. Ketchum, Assoc. M. Am. Soc. C. E., 
from his experience, for spans up to 150 feet: 



45 \ 5 V A/' 



where W = the weight of truss per square foot of hori- 
zontal projection; P=the capacity of the truss in pounds 
per square foot of horizontal projection; L = the span of 



3° 



GRAPHIC STATICS. 



[Ch. I. 



the truss in feet ; and A = the distance between trusses in 
feet. In general, however, for purposes of determining 
the dead-load stresses, it will be sufficiently accurate to 
use the following approximate table: 



For a Span 
Length in Feet of 


Weight of Truss in 

founds per Square 

Foot of Ground 

Surface Covered. 


20 
40 
60 
80 
100 


2 

3 

4 

5 
6 



Great accuracy cannot be expected from the use of this 
table ; but since, in any event, the greatest stresses are caused 
by the weight of truss covering, by wind and by snow loads, 
a small error in assuming the dead weight of a roof -truss is 
not appreciable. If the error be found appreciable, however, 
after tentative design, the proper corrections must be made. 

Trusses having spans up to 50 or 60 feet usually have 
one end supported on planed plates; trusses of greater 
span usually have one end on rollers or on rockers. These 
precautions are necessary on account of temperature 
changes. In the case of the roller or rocker end it is usual 
to assume the reaction at that point perpendicular to the 
plane on which the rollers move, although on account of 
friction this condition may not be rigorously exact. 

The pitch of the roof is usually given as a ratio of the 
centre height divided by the span length; it may vary 
between the limits of 1/2 and 1/5 ; its more usual value is 
1/4. 

Snow Load. 

The snow load carried by a structure depends not only 
on the latitude of its location, but also on the pitch of the 



Art. 13] STRESSES IN ROOF-TRUSSES. 31 

roof. It is ordinarily specified as a load in poionds per 
square foot of horizontal projection. In the vicinity of 
New York a value of 20 pounds per square foot for flat 
roofs is usually taken, and this is decreased for roofs with 
greater pitch. If the pitch is 60° or more, no snow loads 
need be taken, although a minimum weight of 10 pounds per 
square foot, due to sleet, is sometimes specified. The highest 
value for flat roofs in cold climates is 30 pounds per square 
foot, .while for southern latitudes the snow load disappears. 

Wind Load. 

In treating horizontal wind loads on roof-trusses, the 
component normal to the slope of the roof is usually taken, 
the component parallel to the slope of the roof being 
neglected. The intensity of horizontal wind pressure on 
a vertical surface is generally specified at 30 pounds per 
square foot.* 

The normal pressure on a roof due to horizontal wind 
force is not usually found by its simple resolution into 
two rectangular components, but is determined by means 
of an empiric formula based upon experimental work. 
Two such formulae are in common use, one by Hutton, 
given in eq. (i), and the other by Duchemin, given in eq. (2). 

P„=Psin a'-^*^<=°^"-S (i) 

2 sin a 
" i+sin^a' ^^^ 

where P„ represents the normal component, P the horizontal 
force, and a. the angle between the roof surface and a 
horizontal plane. 

* For a more detailed statement concerning wind pressures, see Supplementary 
Report by Capt. W. H. Bixhy in the Report of the Board of Engineer Officers, 
as to Maximum Span Practicable for Suspension Bridges, Washington, 1894. 



32 



GRAPHIC STATICS. 



[Ch. I. 



The following table furnishes the values of the ratios 
of P;; to P for the various slopes of roof indicated: 



VALUES OF 


P» 
P' 


Slope of 
Roof in 
Degrees. 


Duchemin. 


Hutton. 





0.00 


0.00 


lO 

20 
30 


0-34 
0.61 
0.80 


0.24 
0.46 
0.66 


40 


0.91 


0.84 


50 
60 


0.97 
0.99 


0-9S 
1 .00 



Roof Covering. 

The roof covering rests upon longitudinal purlins carried 
by the truss, usually at its panel points. If the purlins 
are not placed at the apices or panel points, bending is 
caused in the chord members. Such stress conditions 
should ordinarily be avoided for simplicity. 

The amount of load carried to any panel point is imme- 
diately determined by noting the area of surface and the 
load per square foot carried by each purlin. This area, or 
its horizontal and vertical projection determines also the 
snow load and the wind load carried at any panel point. 

The weight of roof covering varies with the different 
materials of which it is composed, but it may be closely 
estimated in advance. 

The weights of roof coverings may be approximately 
assumed as follows : 

Iron Sheets. — The weight of iron or steel sheeting 
depends on the gauge thickness and whether the sheets are 
flat or corrugated. The exact weight may always be found 
in manufacturers' handbooks ; it varies from i to 3 pounds 
per square foot for the thicknesses ordinarily employed. 



Art 14] STRESSES IN ROOF-TRUSSES. 33 

Felt, Pitch, and Slag or Gravel Roofing. — This combina- 
tion of materials may weigh from 8 to 10 pounds per square 
foot, depending generally on the number of thicknesses of 
felt. 

Slate. — Slate usually weighs from 7 to 9 pounds per 
square foot of roof. 

Tile. — Terra-cotta tile i inch thick weighs about 6 
poimds per square foot. 

Tin. — Tin, without sheathing, weighs from i to i^- 
poiinds per square foot. 

None of these weights includes weights of purlins or 
sheathing. 

Wooden Coverings. — The weight of wooden roof cover- 
ings may be estimated by assuming the weight of wood 
at 4 pounds per foot B. M. 

The exact weight of roof covering, including sheathing, 
purlins, bracing, gutters, ventilators, etc., must be calcu- 
lated for each individual problem. In a similar way, all 
suspended weights must be determined before the stresses 
can be computed. 

The total loads carried by any truss having then been 
estimated, the determination of the stresses in the members 
of the structure is the next procedure. This determina- 
tion may be made by combining the various classes of 
loads and proceeding with resultants or by treating each 
class separately and subsequently combining the stresses 
so found for each member. The method of procedure 
will be obvious in each case. 



Art. 14. — Stresses in a Roof -truss, Both Ends Fastened. 

A common form of roof -truss is illustrated in Fig. 32, 
in which the span is 40 feet, the rise of the peak 10 feet, 
and the distance between trusses 12 feet. The roof 
covering for this truss is estimated to weigh 10 pounds 



34 



graphjc statics. 



[Ch. I. 



per square foot, and since the total exposed surface between 
two neighboring trusses is 45 feet long by 12 feet wide, 
the total roof covering weighs 45X12X10 = 5400 pounds. 
The dead weight of the truss itself has been estimated at 
3 pounds per square foot of horizontal projection. The 
area of horizontal projection is 40X12=480 square feet; 



\[ J. A--H^ 



>^'"\ 




Fig. 32. 

the dead weight of one truss is, therefore, 1440 pounds, and 
the total dead weight 5400 + 1440 = 6840 pounds. This may 
be supposed equally divided among the upper panel points, 
the end points carrying but half a panel load. Each panel 
load is, therefore, 855 pounds. It is obvious that the end 
panel loads, if vertical, need not be considered as a factor 
in causing stresses in the structure, since these are carried 
directly by the abutments. Vertical end panel loads will, 
therefore, net be considered in the cases to be discussed. 

The dead-load stress diagram is shown in Fig. 2>2)'' its 
construction is as follows: 

Since the load is symmetrical, each reaction is equal 
to half the total load on a truss. The line ah is therefore 
drawn vertically upward and equal by scale to the left- 
hand reaction. The stresses in the two members meeting 
at the left-hand abutraent may at once be found, as there 
are only two unknown quantities. The lines bf and fa are, 
therefore, drawn parallel to the corresponding lines in 
Fig. 32 through the points h and a to their intersection at 
j, thus determining the stresses hf and fa. The circular 



Art. 14.] 



STRESSES IN ROOF-TRUSSES. 



35 



arrow in the upper part of Fig. 32 indicates the direction 
in which the forces are read about any panel point, and 




Fig- 33- 

this order must be carefully observed. By transferring 
the direction of bf from Fig. 33 to Fig. 32, it will be seen 
that the stress acts toward the panel point, indicating com- 
pression. Similarly, fa acts away from the panel point, 
indicating tension. Hereafter the explanation determining 
the sign of the stress in any member need not be given. 

Panel point 2 is next to be considered. With the dis- 
tribution of load assumed there will be no stress fg, and 
fa must evidently be equal to ga. 

Panel point 3 is next in order. The lines ch and hg are 
drawn parallel to their respective lines of action in Fig. 32. 
The remaining construction is precisely similar in character 
and needs no further explanation for the half of truss 
under consideration. 

Fig- 33 represents the stresses for the left half of the 
truss only ; since the structure is symmetrical the members 
in the right half carry identically the same stresses as 
the corresponding left-hand members. The values of the 
stresses as scaled from Fig. t,7, are given in Table I. The 



36 GRAPHIC STATICS. [Ch. I. 

positive sign indicates tensile stress, and the negative sign 
compression. 

A separate diagram for snow loads over the entire roof 
need not be drawn for roof -trusses of this character, since 
it is at once evident that the snow-load stresses may be 
found as a simple ratio of the dead-load stresses. If the 
snow load be taken at 20 pounds per square foot of hori- 
zontal projection, the ratio of the snow-load stresses to 
those due to the dead load will be 

20(40X12) 

6840 ='-4- 

The values of the stresses obtained from this ratio are 
given in Table I. 

The intensity of the wind load, which acts on one side 
of the roof only, has been taken at 30 pounds per square 
foot of vertical surface. By means of Hutton's formula, 
the normal component for this truss will be 30X. 59 = 17 
pounds per square foot, since the angle of inclination of 
the roof surface is 26° 30'. The distance between upper 
panel points is 5.63 feet. Each panel load will therefore 
be 5.63X12X17 = 1150 pounds, acting normally to the 
roof surface. The wind panel loads at the peak and at the 
abutment will, of course, be only half as great, or 575 
pounds. 

The wind stresses in the truss will be found on the 
assumption that the reactions at the two ends of the truss 
are parallel to each other and to the direction of the wind 
loads. In order to determine the amounts of these re- 
actions, the funicular polygon m.ust be employed. The 
loads are first laid down to scale, as shown in Fig. 34. 
Any pole, as 0, is chosen and the rays i, 2 ... 6 drawn 
and transferred to Fig. 32, the polygon being started at 
the left-hand reaction point with ray 2 . Since this polygon 



Art. 14.] 



STRESSES IN ROOF-TRUSSES. 



37 



must close for equilibrium, the missing ray 7 lying between 
the lines of action of the reactions may at once be drawn. 
If this ray be transferred to Fig. 34, it will determine at 
once the point a and give the reactions r'a and ar. The 
stresses in the structure for the wind load may then be 




Scale in Pounds 

I 

Fig. 34. 



found exactly as in the case of the dead load, the left-hand 
reaction point being again the first point treated. The 
force polygon, Fig. 34, thus being constructed precisely 
as was Fig. 33, requires no further explanation. 

Attention is called to the fact that the stresses in all 
the web members of the right half of the truss are zero for 
the direction of the wind shown ; but since the wind may 
be taken to blow in either direction, the stresses in the 
right-hand members will be found to be the same as in 
the left-hand members, if the direction of the wind be re- 
versed. The values of the wind stresses are given in 
Table L The final stresses for which the members of this 
truss must be designed are those found by combining the 
results due to dead, snow, and wind loads ; they are shown 
in the last column of the table. Since no member suffers 



38 



GRAPHIC STATICS. 



[Ch. I. 



Teversal of stress, the sum of all the stresses, viz., for dead, 
'for snow, and for wind loads, determines the maximum 
' stress existing in each member. 

■ ' If, however, in the judgment of the designer, the wind 
stresses and the snow stresses might never occur at the same 
time, only the sum of those stresses which might act simul- 
taneously should be taken. 



Table I. 





Member. 


Dead-load 


Snow-load 


Wind-load 


Maximum 






Stress. 


Stress. 


Stress. 


Stress. 




BF 


— 10,500 


— 14,700 


— 5,200 


-30.400 


Upper 


CH 


- S.460 


- 7.650 


-4,350 


— 17,460 


chord 


DJ 


— 4,600 


- 6,450 


-3.450 


— 14,500 




EL 


- 3.650 


— 5,100 


— 2,600 


-11.350 




AF 


+ 5.75° 


+ 8,050 


-1- 5.800 


+ 19,600 


Lower 


AG 


+ S.750 


+ 8,050 


-t- 5,800 


+ 19,600 


chord 


AI 


+ 4,950 


+ 6,930 


-t- 4,500 


+ 16,380 




AK 


+ 4,100 


+ S.750 


+ 3.200 


+ 13.050 




FG 
















GH 


— 900 


— 1,260 


— 1,400 


- 3,560 


Web 


HI 


+ 420 


+ 590 


+ 650 


+ 1,660 


members 


IJ 


— 1,200 


— 1,700 


-1,850 


- 4.750 




JK 


+ 850 


+ 1,200 


+ 1,300 


+ 3.350 




KL 


— 1,500 


— 2,100 


— 2,300 


— 5.900 


•■■' 


LU 


— 2,480 


- 3.480 


-1.95° 


- 7.910 



Art. 15. — Stresses in a Roof-truss, One End on Rollers. 

\\>: . 

■' ' ' Let the truss of the previous article have one end on 
'(roUets and let the wind stresses be determined. 

^1' ' It will be unnecessary to repeat the determination of 
:ithe stresses caused by vertical loads, since they are in no 

way I affected by the roller end. The values shown in Table 
• ill. for the dead and snow loads are the same as those in 

T^ble I. 

■1!.' The lines of action and the amounts of the reactions 
'ifor 'the wind loads must first be found. In determining 
■■tli©'CeaGtions> the problem is simply the general case of a 



Art. 15 



STRESSES IN ROOF-TRUSSES. 



39 



series of forces in equilibrium, all known except two, of. 
which the point of application of the one and the line of, 
action and direction of the other are known. Assuming the 
direction of the wind from left to right and the roller end 
at R', the funicular polygon (Fig. 32) 2, 3 ... 6 is drawn as 
shown, the only precaution being that the polygon must be 
started at the point of application of the unknown force!/?. 
Th - pole chose.i, and the rays 2, 3 ... 6 used, are 
shown in Fig. 35. The ray 8 is found (Fig. 32) by draw- 
ing it from y so as to close the figure. The direction of ray 8 




Scale in Pounds '' 

Fig. 35. — Stress Diagram; Right End on Rollers. V 

is then transferred- to Fig. 35, where it defines the point a,. 
which determines at once the right-hand reaction r'a and the 
left-hand reaction ar. The further construction of the stress 
diagram is precisely the same as was followed for Fig. 34. 

There now remains the construction of a diagram for 
the wind blowing from left to right, and with the same end 
on rollers. But this problem may be treated more advan- 
tageously by retaining, as in Fig. 35, the direction of the 
wind from left to right, and transposing the roller end from 
the right end to the left end. The stresses found from 
such a diagram (Fig. 36) must, however, apply to the 
members placed in a symmetrical position about the centre 



40 



GRAPHIC STATICS. 



[Ch. I. 



line of the truss ; that is, the letters designating the stresses 
in Fig. 36 are all primed letters. This method of treatment 
evidently does not apply to unsymmetrical trusses. 

The reactions for this case may be found most simply 
in the following manner: It is evident that the vertical 
components of the two reactions are the same, no matter 
which end of the truss is on rollers; therefore, in Fig. 36, r'a 




Scale in Pounds 
Fig. 36. — Stress Diagram; Left End on Rollers. 



may at once be drawn from Fig. 35, since it is merely the 
vertical component of ra there determined. It is only neces- 
sary then to connect the points r and a (Fig. 36) to obtain 
the reaction ra. After these reactions are determined the 
stress diagram is to be constructed precisely as before. 

The final stresses in the structure for all possible loads 
are shown in Table II. The final column shows the range 
of stress to which each member may be subjected and for 
which it must be designed; it is to be observed that the 
dead-load stress must always appear in that range, the 
other stresses being included only as the judgment of the 
designer may indicate. It is evident also that in the range 
of stress for any member there must appear but one 
value of the wind stress, for the wind cannot act simul- 
taneously in both directions. 



Art. 15.] 


STRESSES 


IN ROOF-TRUSSES. 


4-1 




Table II. 








Member. 


Dead-load 


Snow-load 


Wind-load 
Stress; 


Wind-load 

Stress; 

Wind 

Right to 

Left. 


Range o£ 




Stress. 


Stress. 


Wind Left 
to Right. 


Stress. 




' BF=B'F' 


— 10,500 


-14,700 


-5,200 


-5,200 


1 

i -10,500 
1 -30,400 


Upper 


CH = C'H' 


- 5-460 


- 7.650 


-4,350 


-4,350 


j - 5,460 
( — 17,460 


chord 


DJ = D'J' 


— 4,600 


- 6,450 


-3-450 


-3,450 


( — 4,600 
I -14,500 




EL = E'L' 


- 3-650 


— 5,100 


— 2,600 


— 2,600 


j - 3,650 
/ -11,350 




AF 


+ 5-750 


+ 8,050 


+ 6,500 


+ 400 


J + 5,750 
( +20,300 




AG 


+ 5,750 


+ 8,050 


+ 6,500 


+ 400 


j + 5,750 
1 +20,300 




AI 


+ 4-950 


+ 6,930 


+ 5,200 


+ 400 


1 + 4,950 
1 +17,080 


Lower 


AK 


+ 4,100 


+ 5,750 


+ 3,900 


+ 400 


J + 4,100 
1 +13,750 


chord 


AK' 


+ 4,100 


+ 5,750 


+ 2,600 


+ 1,700 


j + 4,100 
\ + 12,450 




AF 


+ 4,950 


+ 6,930 


+ 2,600 


+ 3,000 


/ + 4,950 
\ + 14,880 




AG' 


+ S-750 


+ 8,050 


+ 2,600 


+ 4,300 


j + 5,750 
) +18,100 




AF' 


+ S-750 


+ 8,050 


+ 2,600 


+ 4,300 


j+ 5-750 
i +18,100 




■ FG 



















GH 


— 900 


— 1,260 


— 1,400 





J — 900 
( - 3-560 




HI 


+ 420 


+ 590 


+ 650 





j + 420 
\ + 1,660 




IJ 


— 1,200 


— 1,700 


-1,850 





j — 1,200 
\ - 4-750 




JK 


+ 850 


+ 1,200 


+ 1,300 





j + 850 


Web 


KL 










\ + 3,350 


members 


- 1,500 


— 2,100 


— 2,300 





J - 1-500 














/ — 5,900 




LU 


— 2,480 


- 3-480 


-1,950 


-1-950 


- 2,480 
/ - 7,910 




K'U 


— 1,500 


— 2,100 





— 2,300 


Same as KL 




J'K' 


+ 850 


+ 1,200 





+ 1,300 


" " JK 




I'J' 


— 1,200 


— 1,700 





-1,850 


" " IJ 




H'l' 


+ 420 


+ 590 





+ 650 


" "HI 




G'H' 


— 900 


— 1,260 





— 1,400 


" " GH 




F'G' 














" " FG 



Comparison of Tables I and II 
on rollers no stresses differ from 



shows that for one end 
those foiind with both 



42 GRAPHIC STATICS. [Ch. I. 

ends fastened, except in the lower chord. Table II fur- 
nishes lower values for the lower chord members near the 
fixed point of support, and higher values at the roller end. 
This statement as to variations in stress for the two methods 
of end support is not a general one, although it applies 
in this case, where the entire lower chord is in one straight 
line. 

Connterhraces. 

If the range of stress to which a web member is sub- 
jected includes values of both tension and compression, 
that member must be designed to resist both; it is then 
said to be counterbraced. If such a member is constructed 
that it can resist tension only, provision for the com- 
pressive stress may be made by inserting in the same panel, 
and crossing the original member, an additional member 
known as a counterbrace. It will be found that this 
member will be stressed in tension when the compressive 
stress in the original member causes the latter to be useless ; 
the stresses in the members of the structure affected by this 
new condition may then be found by assuming that the 
counterbrace is the main member, and that the original 
member does not exist. 

Art. 1 6. — Fink Roof-truss. 

Fig. 37 represents an application of the P'ink truss to 
roof construction. It presents a slightly more complex 
problem than the truss illustrated in the previous articles. 

The following data are assumed: 

Span = 6o feet ; 

Rise of peak = 15 feet ; 

Distance between trusses = 14 feet. 

The roof covering, including purlins and wind bracings, is 
taken at 1 2 pounds per square foot of inclined surface, and 



Art. i6.] 



FINK ROOF-TRUSS. 



43 



the dead weight of one truss at 3.8 pounds per square foot 
of horizontal projection. A panel load will then be found 
to be 1175+400 = 1575 pounds. The stress diagram 
(Fig. 38) is again begtm by drawing mh vertically upward 
as the left-hand reaction and determining the stresses in 
hf and fm ; the direction of rotation about any point being in- 
dicated by the circular arrow (Fig. 37.) Panel points 2 and 3 






1 


a 


1 
1 


E 
4 


2,^ 


.,-f- 

■^ 


D 


^r 


<i 


c 


-^ 


^ \ 


/ ' 




Span = 60 feet. 



Rise = 15 feet. 



Distance between trusses = 14 feet. 



Dead Weights Assumed. 
Roof covering including purlins and wind-bracing = 1 2 pounds per sq. ft. stirface. 
Truss = 3.8 pounds persq. ft. horizontal projection. 
Snow load = ,^o lbs. per sq. ft. horizontal projection. 
Wind load =40 lbs. per sq. ft. vertical projection. 

Fig. 37. 

may then be treated in precisely the way already explained, 
but it will be found that at both panel points 4 and 5 there 
will be three unknown quantities instead of two, atid that 
recourse must be had to some other method than the use 
of the simple force diagram. A section is, therefore, 
passed through the truss, cutting, as shown, the members 
EK, KL, and LM. The stresses in the members cut hold 
in equilibrium all of the external forces on one or th.e other 
side of the section. In this case the left-hand forces will 
be considered. 

The stresses in these three members may be determined 
by means of the general principles already explained, for 



44 



GRAPHIC STATICS. 



[Ch. I. 



the problem is simply that of a set of forces in equilibrium, 
all known except three, the lines of action of the latter 
being given. The funicular polygon is, therefore, drawn, 
the only precaution being that the polygon Fig. 37 must 
start at the intersection of two of the unknown forces. 
In this case the peak was chosen as the starting point, 
being the intersection of the members EK and KL. The 
rays 4, 3, 2, i, and 5 were then drawn. As the forces are m 
equilibrium the polygon must close, and the closing ray 6 
may then be drawn and transferred from Fig. 37 to Fig. 
38, \\'here it immediately defines the point m and determines 




Fig. 38. 



the stress in the member ML. Knowing then the stress 
in LM, panel point 5 may be treated in the usual manner, 
as the stresses in the members HI and IL are the only 
unknown forces. Panel point 4 may next be treated in 
the usual manner and all the stresses be determined with- 
out further explanation. 

A more simple solution of this problem is possible by 
determining the stress in the member LM by taking moments 
about the peak and then inserting its value in the force 



Art i6.] hlNK ROOF-TRUSS. 45 

diagram Fig. 38. The graphic solution for the remaining 
stresses can then proceed in the usual way. The moment 
equation for the peak as a centre would then be as follows : 

MBXzo-BCX22\-CDXiS-DEXlh-LMXiS=o, 

or 

55 10 X 30 - 1 575(22^ + 15 + 7*) -LilfX 15=0, 

and the stress in the member LM would be 

LM = + 6300 pounds. 

This, of course, should check the value of the stress found 
graphically. 

It is proper to explain also one other method of solution 
of the Fink truss, as illustrated in Figs. 39 and 40. It is 
the method of substitution of diagonals, and is shown 
applied to the right half of the truss. If the members 
K' J' and /'/' be removed from the structure and be re- 
placed temporarily by the member YX, the stress in the 
member LAI may be found by the methods of the ordinary 
force polygon, the panel points i, 2, 3, 4, 5, and 6 being 
taken in the order named, since at no panel point will 
there be more than two unknown forces. This substitution 
of the diagonal YX does not affect in any way the stress 
in the member LM, for the stress in that member cannot 
be affected by changes in any other panel of the character 
indicated. In fact, the portion of the truss defined by 
panel points 7, 6, and i might be replaced by a^iy rigid 
mass. Having, therefore, found the stress in the member 
LM by this method of substitution, the original web 
members K'J' and /'/' may again be replaced and their 
stresses found in the usual manner, without further diffi- 



46 GRAPHIC SI A'l !€::>. ^Ch. I. 

culty. No further use is made of the dotted hne xy in 
Fig. 4°- 



Fig. 39. 




^cale in. Pounds 



Fig. 40. 

The treatment of the wind stresses in a Fink truss need 
not be considered here in detail, as the methods to be used 
are the same as those employed in the preceding articles. 



Art. 17. — Unsymmetrical Trusses. 

The treatment of unsymmetrical trusses is precisely 
the same as for those that are symmetrical. Proper con- 
sideration should, however, be accorded to the determina- 
tion of the panel loads, since these are in general no longer 
equal, and it will usually be convenient to determine the 



Art. 17. J 



UNSYMMETRIC/IL TRUSSES. 



47 



reactions of the truss by means of the funicular polygon. 
In Fig. 41, which represents an unsymmetrical truss carry- 
ing at the upper panel points the weights shown, the 
reactions es and sa were determined by means of the 
lower funicular polygon i, 2, 3, 4, 5, 6, the directions of these 




Fig. 41. 



Scale in Feet 

Span=60 Feet 

Rise of Peak =20 Feet 

Distance of Peak ; 

From Right Abutment =18 Feet 

Distance between Trasses =12 Feet 




Scale in Pounds 



Fig. 42. 



rays being obtained from Fig. 42. The transference of 
ray 6 from Fig. 41 to Fig. 42 then fixes the point 5 which 
determines the amounts of the two reactions. The finding 
of the stresses presents no difficulties until panel point 4 
or 5 is reached, when one of the constructions of the previous 



48 GR/iPHIC STATICS. [Ch. I. 

article must be employed. In this case the funicular 
polygon becomes 4', 3', 2', i', 6', the point of beginning 
being chosen at the peak. The ray 7 becomes the closing 
line, and being transferred to Fig. 42 determines the position 
of the point /. The remaining stresses may then be found 
in the usual manner. 

The preceding constructions are all that are necessary 
to determine the fixed or dead-load stresses in any simply 
supported structure, and they are immediately available 
for finding the dead-load stresses in railroad or highway 
bridge-trusses. They are also applicable for finding the 
dead-load stresses in the open arms of swing bridges. 
These constructions are equally applicable to the dead- 
load stresses in both cantilever bridges and three-hinged 
arches, but the treatment of the fixed-load stresses in such 
structures will be considered at the same time with the 
moving loads. 



Art. 18. — Bending of Supporting Columns of Roofs. 

The roof-trusses of the preceding articles have been 
treated as resting on supports or walls capable of resisting 
the horizontal forces acting against them, but the stresses 
in which do not affect the stresses in the truss members. 
If, however, the roof -truss be a part of the transverse bent 
in the building (Fig. 43) connected to the vertical columns 
by knee-brackets, then the bending stresses produced in 
the columns cause additional stresses in the members of 
the truss. 

The dead loads and snow loads resting on the roof are 
vertical forces that cause no bending stresses in the columns 
and no stresses in the knee-brackets, provided the deforma- 
tion of the truss itself be neglected. If the deflections of 
the truss are considerable, however, additional stresses will 



Art. i8.j BENDING OF SUPPORTING COLUMNS OF ROOFS. 



49 



be produced in the knee-brackets and columns, and also 
in the members of the truss; generally such deflections 
are not considered. It is evident, then, that the methods 
of the previous articles when treating vertical loads apply 
also to trusses mounted on columns. Non-vertical loads 



< /77/, 




a ^ 



V 



I 
I 
I 
j, a' 






F^G. 43. — Posts Hinged at Top and Base. 



require further consideration. Such loads may include 
not only wind loads, but loads caused by shafting, hoists, 
cranes, etc. In this article horizontal wind loads only 
will be considered, but the treatment will be general, so 
that it may be applied to loads of any character which 
may be resolved into vertical and horizontal components. 
The wind forces acting against the vertical side of the 
building and causing flexure in the posts must be included 
in the loading. Since that surface is usually much greater 
than that exposed by the roof -truss itself, it is usual in the 
consideration of this type of roof-truss to treat the wind 
load on the truss as horizontal, instead of normal to the 
surface, as previously, but the method is equally applicable 
to normal loads. 



50 GRAPHIC STATICS. [Ch. I. 

Fig. 43 illustrates a roof -truss mounted on the posts ac 
and a'c' and connected to them by the knee-brackets bd 
and b'd'. It is generally assumed that the upper ends c 
and c' of the posts or columns are hinged to the truss so 
that the latter is free to turn about those points ; this 
involves the condition that there is no bending moment 
in the posts at those points. Such a condition may never 
be realized in practice, but it is on the side of safety. The 
fastening of the bases a and a' of the columns admits, 
however, of two distinct methods of treatment: 

Case I. The columns may be considered hinged at a 
and a' and free to turn about those points. 

Case II. The columns may be considered firmly fastened 
at a and a'. In that case the anchorage must provide for 
the bending moments existing at those points. 

In practice ■ it is doubtful if the conditions of either 
Case I or Case II are ever exactly fulfilled. The true 
condition lies probably between the limits indicated, but 
it will always be found on the side of safety, as far as the 
truss or post stresses are concerned, to treat all problems 
by the methods of Case I. It is to be observed that Case I 
requires no anchorage in the base to resist bending, while 
Case II requires an anchorage. 

Case I. 

The problem is most easily solved by three distinct 
operations: (i) Determining the forces causing bending 
stresses in the columns; (2) transferring the loads created 
by these bending stresses to the panel points of the framed 
structure, and (3) finding the stresses in the members of 
the framed structure for these loads and all other external 
loads acting upon it. 

The loads to, be treated are shown in Fig. 43 as Pi, P2, 
P3, and Pi ; under certain conditions it will prove convenient 



Art. i8.] BENDING OF SUPPORTING COLUMNS UF ROOFS. 51 

to employ their resultant R acting at a distance / above 
the surface. It will first be necessary to find the rect- 
angular components H, V, Hi, and Vi of the reactions 
acting at a and a', the bases of the posts. In the following 
treatment H will always be taken equal to Hi ; it is perhaps 
possible, by an analysis based on the deflections of the 
structure, to obtain more accurate values, but such refine- 
ment is unnecessary. In the present case, therefore, 



H = Hi=~ (i) 



Taking moments of all the external forces about a' , 

R.f-V-m^o or V = ^=^-Vi. . . (2) 

The structure may now be divided into the three parts 
shown in Figs. 44, 45, and 46. Figs. 44 and 46 represent 
the forces acting on the posts ac and a'c' respectively, while 
Fig. 45 shows all the panel loads acting on the statically 
determinate framework there shown. The forces causing 
flexure in the -posts are also loads acting on the truss, 
but they are exactly reversed in direction. This is evident, 
for the forces Qi, Q2, Qz, and Q4 are simply used as tem- 
porary devices to indicate forces acting on the posts and 
must be eliminated frora the problem by considering them 
as loads causing stresses in the triangular framework. 
Fig. 45 therefore shows, as loads at the panel points, the 
forces Pi, P2, Pz, and P4 and also the forces Qi, Q2, Qz, 
and Qa, every one of the latter forces being exactly equal 
but opposite in direction to the similarly lettered forces 
of Figs. 44 and 46. 

Since the conditions of the problem state that the 



52 



GRAPHIC STATICS. 



[Ch. L 



posts are hinged at the bases, each may be treated as a 
simply supported beam carrying a concentrated load Q2OT 




8 gfa 

Ill 
Fig. 44- 



All the Forces acting on 
Triangular Frame Work. 



Fig. 45- 








^( 




s 












Fig. 46. Fig. 47. Fig. 



Qi at 6 or 6', the points of attachment of the knee-brackets. 
Treating post ac and taking moments about b, 

H-g = Q,{h~g). 



Qi = 



h-g' 



(3) 



and from the relation that IH = o, 



Q2=//+Qi. 



(4) 



Eqs. (3) and (4) furnish the values of the forces Qi 
and Q2, and in precisely the same way there may be found 
the values of 0^ and Q4. The shear and bending moment 
existing at every point in the post a'c' are represented 
graphically in Figs. 47 and 48. The column section must 



Art. i8.J BENDING Oh SUPPORTING COLUMNS OF ROOFS. 53 

be capable of carrying these shearing and bending stresses 
in addition to its direct stress of tension or compression.* 

The second step in the analysis is to transfer in the 
proper manner to Fig. 45 the forces Q\, O2, Qs, ^nd O4 and 
to unite them with the loads P wheneA'er they act at the 
same panel points. For instance, at panel point c the 
load is —Qi+Ps, whereas at c' it is only — O3; similarly 
at b the load is —02 + Pi, wdiereas at b' it is only —Q4. 
It is also necessary to insert as loads on the structure the 
vertical forces T and —Vi at the points b and b', which 
are also the values of the direct stresses in the posts ab 
and a'b', for they are the vertical components of the reac- 
tions transferred directly from the bases a and a'. 

The stresses in the members of the framework of Fig. 
45 may then be found at once by a single stress diagram 
which it is not necessary to reproduce. It will be found 
that the portions be and b'c' of the posts sustain different 
sti'esses from the lower portions ab and a'b' of the same 
members. 

After having obtained the direct stresses in the members 
6c and b'c', these must be united in the proper manner 
with the bending stresses previously found. It is seen 
that the preceding treatment is perfectly general and may 
be applied to other than horizontal forces. 

Case II. 

Fig. 49 illustrates the second method of treatment 
in which the posts are assumed to be rigidly connected 
at the bases a and a'. In consequence of this rigid con- 
nection each post acts similarly to a beam (Fig. 50) which. 



* For the design of members subjected to combined stresses, such as flexure 
and tension or flexure and compression, see Burr's "Resistance of Materials," 
p. 164. 



54 



GRAPHIC STATICS. 



[Ch, I. 



is rigidly fastened at one end a, simply supported at the 
other end c, and which carries a weight at h acting at right 
angles to the axis of the beam. The post is represented 



i ^H 







Fig. 49. — Posts Hinged at Top and Fixed at Base. Points of Contraflexure 

at e and e' . 



in Fig. 50 as a horizontal beam, and it contains at some point 

in its length a point of contraflexure, or point of no bending, 

as at e. If it be assumed 
that the points h and c deflect 
equal amounts hh" and cc" , 
the following analysis will 
locate the point of contra- 
flexure. 

Fig. 50. The bending moment to 

which the portion ah of the 

beam is subjected at any point distant x from a may be 

represented as follows: 




M,, = Qr{h-x)-Q^{g-x)=EI 



dx^' 



(5) 



Art. i8.] BENDING OF SLTPORTING COLUMNS OF ROOFS. 55 

The last member in eq. (5) is derived from the theory 

of flexure, in which the external bending moment at any 

d-y 
section is equal to EI j—„ where E is the coefficient of 

elasticity, / is the moment of inertia of the cross-section 
of the beam, and y and .x are the respective coordinates of 
any point. Integrating eq. (5), the tangent of inclination 
at any point of the length ab, multiplied by EI, will be 

£/g=Q:(te-f)-0.(gx-f) + (C=o).. . (6) 

C=o, since -r=o when x=o. 
ax 

For the portion he of the span the mometat is 

M,, = Qi(/.-:.)=£/g (7) 

Integrating as before, 

£/|-Q.(*--f)+G (8) 

dy 
Ci may be evaluated, for El-r^ in eq. (8) has the same value 

as in eq. (6) at the point b where x=g. 

Therefore Ci = -^^ 

2 

and eq. (8) becomes 



56 GRAPHIC STATICS. [Ch. I. 

Integrating again, 

y in eq. (lo) represents the deflection of any point of the 
beam between b and c. When x=g or h, the deflections 
are equal, according to the premises of the problem ; that is, 



K¥-?) 



Qi(— --)--^ + C2 = Qi(y-y) —+C,. (II) 



Cancelling C2 from each member, the following relation 
may be obtained: 



Since the moment of the external forces is zero about 
the point of contraflexure, 

Qiih-k)=Q2(g-k). . . . \ . (13) 

Qi g~k 

Q2^h^ ^^4) 

Subtracting eq. (14) from eq. (12), and solving for k, there 
is obtained 

z, g/2h + g\ 

^^Ah + ^g) ^'5^ 

Table I furnishes the values of k in terms of h for 
■various values of g in terms of h. 



Art. i8.] BENDING OF SUPPORTING COLUMNS OF ROOFS. 

Table I. 



57 



When g equals Then k equals 


h 


5og 


.gh 


S^g 




8h 


54^ 




jh 


56^ 




6h 


59^ 




5h 


62g 




Ah 


(>7g 




3h 


73.? 


2h 


75g 



Since g is seldom less than .5/1, and since moreover the 
effect of the vertical loading on the column has not been 
included in the treatment, Table I shows that it will prove 
sufficiently accurate to take the point of contraflexure 
midway between a and b. 

Having determined the position of the points of contra- 
flexure, e and e', in Figs. 51 and 53, the forces causing flexure 
in the post may at once be found. The windward post 

only will be treated. 

70 

As in Case I, H = Hi=—. Since the moment about e 

2 

equals zero, 

also, 
therefore 



Qi(h-k)==Q2(g-k); 
Qi=Q2-H from IH=o; 
H{h-k) 



02 = 



h-g 



(16) 



or precisely the same value as if the post had a length ec 
and the force H were acting at e, the latter point being 
considered a hinge support ; that is, as far as the bending 
in ec is concerned, the force H oX a might be replaced by 



58 



GRAPHIC STATICS. 



[Ch. I. 



an equal force H2 acting at e. The moment at the base a 
is then H2-k=H-k. In the same way a force Hz=H\ 
may be placed at e' . 

The vertical components of the reactions may now be 
fotind. Taking moments of the external forces about a, 

R-f-H2-k-H3-k~V-m=o. . . . (17) 

But H2 + H3=R; therefore eq. (17) may take the form 

R{f-k)=V-m=-Vim. .... (18) 

The values of V and Fi are therefore precisely the same 
as if the posts were hinged at the points e and e', and it is 




^ TS S 
o] & to 

S 13 .g 



" m g 
FPU g 
'"§ "^ r^ 




=Hi (g-fc) 



M=Hib 



Fig. 51. 



Fig. 52. 



Fig. 53. Fig. 54. Fig. 55. 



seen that Case II is a repetition of Case I if in the former 
the posts be shortened by an amount k. 

Case II therefore requires no further treatment, since 
from this point it is the same as Case I; Figs. 51, 52, 



Art. lii.] BENDING OF SUPPORTING COLUMNS OF ROOFS. 59 

and 53 show the panel loads and other forces acting on the 
members. 

It should be noted, however, that the force P4 in Case II 
has not the same value as in Case I, for in Case II it is 
only the wind load acting on the surface between points 
e and c, whereas it previously was the wind load acting 
between a and c. 

Fig. 54 is the shear diagram for the leeward post. 

The moment diagram (Fig. 55) for the leeward post 
shows a zero moment at the point of contra flexure e', 
and a bending moment Hik at the base, which must be 
cared for by the anchorage. Proper attention should be 
paid to the signs of the bending moments to note on which 
side of the post they cause compression or tension. It is 
also to be noted that the use of H2 and H3 is only a tem- 
porary device to show more clearly the action of the bend- 
ing forces on the post. 



CHAPTER II. 

INFLUENCE LINES FOR SIMPLY SUPPORTED BRIDGE 

TRUSSES. 

An influence line is a line showing the variation in any 
function at any section of a beam or in any member of a 
truss caused by any load moving along such a beam or 
truss. It is clear that such a line can be used to indicate 
the position of the moving load causing the maximum 
shears, moments, reactions, or stresses in any structure, 
and it is for the purpose of indicating and obtaining such 
maxima that influence lines are used. 

In general, influence lines are drawn for a single unit 
load, and unless noted otherwise, it will be assumed that 
they are drawn for such loading only. 

Art. I. — Influence Line for Reaction. 

A reaction influence line is a line showing the variation 
of the reaction of a beam under a 

g i§. X B moving load. If AB (Fig. i) repre- 

_i J _t sents a simple non- continuous beam 

Jc I of length /, and P\ a load moving 

[^^ I over the beam from right to left, its 

D "I'li'i'ii'i'n^^ distance at any instant from the right 

Pjj, j abutment being represented by x, the 

reaction R at the left will be given 

by eq. i, which is the equation of a straight line: 

«=^. (■) 

60 



Art. 2.] INFLUENCE LINE FOR SHEAR. 6l 

If a line KL, parallel to the beam and of equal length, 
be laid off below the beam and an ordinate be erected at 
the left-hand end representing to scale the load Pi, a line, 
ML, connecting the end of this ordinate with the opposite 
end of the base line will be a line expressing eq. i graphi- 
cally. If the load be at the distance x from B and an 
ordinate CD be erected immediately below this point, 
the triangles CDL and MKL will be similar, and therefore 

CD MK ^^ Prx „ 

DL = KL^ or CD=--y=R. 

This equation shows that an ordinate between the line 
ML and KL will represent the reaction at the left-hand 
end of the beam for the load Pi placed at a point imme- 
diately over the ordinate; therefore the line ML is an 
influence line for reaction, since it shows the variation of 
the reaction as the load P\ moves along the beam. 

The inflvience line thus constructed for a load Pi may 
be used for any other load Q, it being necessary, however, 
to multiply the value of any ordinate drawn for Pi by the 
ratio of Q/Pi- This applies to any influence line which 
may hereafter be constructed. 



Art. 2. — Influence Line for Shear. 

An influence line representing the variation of shear at 
any section, CD, in a beam (Fig. 2) as a load crosses the 
span is derived in a similar manner. ^ As a load Pi advances 
towards the section from the right, the shear at any instant 
will be equal to the reaction R, or PiV^, and, as before, 
may be represented graphically by the line LM. 

After passing the section, however, the shear becomes 
equal to the reaction R minus Pi, and is therefore a 



62 INFLUENCE LINES. [Ch II. 

negative quantity. To represent this graphically, ordi- 
nates of a value Pi must be drawn downward from the 
line OM, and the locus of the ends of these ordinates will 
be the line KQ paraUel to the line LOM. The line LOQK 
will then represent the variation of the shear at the section 
CD as the load Pi crosses the span, the shear being positive 

Is, ^ ^ and of an increasing positive value 

■'■jj ip : j^ as the load advances towards the 

„ 7 section, and negative and of a de- 

~^lTnTnTrrr>^ ' creasing negative value as the load 

^^^ n^^ leaves the section. It is seen that 

Q"----^- i the maximum positive shear is found 

"~~---~.J when the load is just to the right of 

^^°- 2- the section in question. In practice 

the load is always placed at the section. 

If two equal loads, Pi, separated by a fixed distance a, 
be employed, the maximum shear will be found when one 
of the loads is at the section, and the value of the shear 
will be the sum of the ordinates erected below the two 
wheel loads. 

If two unequal loads, P2 and P3, separated by a fixed 
distance a, be employed, maximum values of the shear 
will again be found with one load at the section. If the 
influence line be drawn for a load Pi, then for the load Pa- 
placed at the section, both loads P2 and P3 being on the 
beam, the shear will be represented by the algebraic sum 
of the product of the ordinate corresponding to P2 by 
the ratio of P2 to Pi, and of the product of the ordinate 
corresponding to P3 by the ratio of P3 to Pi. 

If the loading advances to the left to such a position 
that the load P3 is at the section, the shear will be repre- 
sented by the algebraic sum of corresponding products. 
Trial alone determines which position causes the greater 
maximum shear at the section ; the second position of the 
loading will usually give the greater maximum shear, if 



Art. j.] reaction INFLUENCE LINES. 63 

P2 be small compared with P3, if the distance a be large, 
or if the section CD be near the left abutment. 

This construction can be used in finding the maximum 
shear at any section of a beam when a series of concen- 
trated loads separated by fixed distances, as in the case 
of a locomotive, is used. The operation in such a case 
is as follows : 

An influence line for a unit weight having been drawn, 
the algebraic sums of the products of ordinates erected 
under the various wheel loads by the actual weight of each 
load must be compared for positions of the loading with 
different wheel loads at the section in question ; the values 
of the quantities so found will indicate not only the posi- 
tion of the loading for maximum shear, but will give the 
value of the shear. This operation is not as tedious as 
it may at first appear, since it is evident that the greatest 
maximum shear usually occurs with the head of the loco- 
motive near the section. Attention is called to the fact 
that should the loading advance so much to the left that 
new loads appear upon the span at the right, such new 
loads must not be neglected. 

Art. 3. — Influence Line for the Reactions of a Series of 
Concentrated Loads. 

The influence lines so far considered have involved in 
their construction the use of only one load; and the use 
of such lines, in the casq of more than one load, has re- 
quired the use of arithmetical calculations. There will 
now be considered the construction of a line such that the 
variation of the reaction of a beam, as well as the reaction 
itself, when a series of concentrated wheel loads passes 
over the beam, may be measured directly from the draw- 
ing. 

In order to simplify the explanation, only three loads, 



64 



INFLUENCE LINES. 



[Crf. IL 



Pi, P2, and Pz, separated by the fixed distances a and h, 
will be used, but the construction is general and may be 
applied with ease to any number of loads. Let the posi- 
tions of the loads be shown in Fig. 3, x representing the 




Fig. 3. 

distance of P3 from the right abutment. Taking rn,oments 
about the right abutment, there will be obtained 



R=j[P,{a-\-h^x)-^P2{h + x)+Pzxl 



(i) 



As before, let KL represent the base line equal in length 
to the span, and let KM represent to scale the loads P\, 
P2, Pz, laid off upwards from K in consecutive order- 
From L lay off to the left, in order and to proper scale, 
the distances a, b, and x. Draw the lines ML, NL, and 
OL. At R erect an ordinate RU to the line LO. From 
U draw a line UV parallel to NL to its intersection with 
an ordinate erected at 5. From V draw a line VW parallel 
to LM to its intersection with an ordinate erected a T. 
Then WT will represent the reaction at the left end of the 
heav.i for the positions of the loads as shown; for WT is 
composed of the three parts, TY, YX, and XW, found 



Art. 3.] RE/1CTION INFLUENCE LINES. 65 

by continuing LU and UV to intersect WT. By con- 
struction, the triangles OKL .and YTL are similar. There- 
fore 

YT OK 

TL "KL 

Substituting the values of those quantities which are 
known, it is found that 

y 1 - ~- ^ - . 

The triangles NOL and XYU are also similar, by construc- 
tion; therefore 

XY NO 
TR ~KL- 

Again, substituting the values of those quantities which are 
known, 

P2{h + x) 



XY = - 



l 



By construction, the triangles MNL and WXV are also 
similar. Therefore 

WX MN ^^,^ P^-x 

By summation, 

YT + XY + WX==j[P,(a + b + x)+P2ib+x)+Psxl 

or the reaction at the left-hand end of the span, as already 
shown by eq. (i). 



66 INFLUENCE LINES. [Ch. II. 

The loads may move any other distance x upon the 
bridge. This involves no change in the figure as drawn. 
To find the reaction for such a change in the position of 
the loading, an ordinate must be erected at a distance 
from the left end of h equal to the new distance x\ the 
intercept on this new ordinate between KL and the line VW 
continued will give the new reaction. The line LUVW 
is thus an influence line for reactions, since it shows the 
variation in the reaction as the loads Pi, P2, and P3 move 
along the bridge. Thus the ordinate immediately below 
Pi always represents the reaction for that position of the 
loading. 

In using a uniform load in this form of construction, 
the uniform load should be treated as a series of con- 
centrated loads spaced as closely as the accuracy of the 
problem may demand. 



Art. 4. — Influence Line for Maximum Shear, for a Series of 
Concentrated Loads. 

It may now be shown that the line constructed in the 
manner described in the preceding article can also be 
used as an influence line to find the maximum shear at 
any section of a beam. Let CD, Fig. 3, be the section of 
the beam under consideration. If Pi, or the first wheel 
load, is at this section, the shear is equal to the reaction 
due to the loads on the beam or to S'V . If the loading 
advances till P2 is at the section, the shear at the section 
becomes equal to the reaction, represented by the ordi- 
nate WT minus Pi. Pi can therefore be laid off down- 
ward from W as WZ. The shear is then represented 
by TZ. If the line V'Z slopes downward to the left, it 
is evident that the shear with wheel load Pi at the section 
is the greater; if the line slopes upward to the left, it is 
evident that the shear is greater with the wheel load P2 



Art. 5.] 



INFLUENCE LINE FOR MOMENTS. 



67 



at the section. Should Po give a greater shear, it is a 
simple matter to test wheel load P3 at the section, or, in 
actual practice, any number of wheel concentrations, as 
in the case of a locomotive. The line LUV'Z . . . 
evidently an influence line for shear at the section CD 



is 



Art. 5. — Influence Line for Moments. 

A line which indicates the variation of bending moment 
at any point in a bearn under a single moving load is a 
moment influence line. Let AB (Fig. 4) represent a beam 




Fig. 4. 

of length /, and let c be the section which divides the beam 
into the two portions /' and I" and about which moments 
are to be found. If a load P" be on the section I" (at 
the point G, distant x' from B, for instance), the moment 
M' about C is the product of the left reaction by the 
length AC ; that is. 



P'^y' 
M'=^-^~-l'=P"m'. 



(i) 



68 INFLUENCE LINES. [Ch. II. 

If the load P" is a i^nit load, eq. (i) takes the form. 

M'=j-l' (2> 

Eq. (2) is the equation of a straight line and may be 
represented graphically as follows: Erect on the base line 
AB at yl a vertical line AE equal in length to /', and 
connect E with B, the opposite end of the span. From 
similar triangles GH: V::x':l; that is, 

Therefore any ordinate between CB and DB represents 

the bending moment at C when the tuiit load is placed 

directly over such ordinate. Similarly, for a unit load 

on /' between A and C and measuring x from A , the moment 

is 

xl'' 
M"=-Y, (3) 

and the line AD, representing eq. (3) graphically, is drawn 
by connecting F, the end of a vertical I" erected at B, 
with A. The line ADB is thus an influence line for mo- 
ments. It is evident from the construction that the comer 
D must lie vertically below the centre of moments. 

If P' is the load on /', the general value of the moment 
M" represented by eq. (3) is 

M"=^l"=P'm (3a) 

If the load is uniform over the entire span and of the 
intensity p, i.e. p being the amotint per linear unit, P" 
becomes p-dxf and P', p-dx. By substituting these values 



Art. 5 ] INFLUENCE LINE FOR MOMENTS. "9 

in eqs. (i) and {^,0) and remembering that differential 
moments then result, 



x' ■ dx' X ■ dx 

dM'=^p-~l' and dM" = p—j~l" . 



The moment M at C then takes the value 



M^Urx'-dx'-V\j^x-dx-l''\=\pl'l". . (4) 



Obviously M has its greatest value at the centre of 

span where V = /" = — and for which 
2 



uJi (5) 



It is also clear from the preceding that the area of 
the triangle ABD multiplied by p represents the value of 
the moment M at C for a uniform load over the entire 
bridge ; its value is 

M = \AB-CD-p (6) 

But CD:AE::BC:AB, therefore 

AE-BC I' I" 
^^~ AB ~ I ■ 

Substituting this value in eq. (6), M = \pl'l", as in eq. (4). 



7° INFLUENCE LINES. [Ch. U 

Art. 6. — Criterion for Maximum Moment at any Section of a 

Beam. 

By means of the influence line of the preceding article, 
the criterion for the position of a series 'of concentrated 
loads producing the maximum moment at any section 
of a beam may be deduced. Let P" be any load on I" 
(Fig. 4), and m' the value of the influence ordinate corre- 
sponding to this load; P' any load on /', and ni the corre- 
sponding influence ordinate; x' the distance of P" from 
B, and x the distance of P' from A ; and then let 71/ be 
the value of the moment at C for any position of the wheel 
loads. If the sign I indicates the summation of terms 
of the same kind, 

M = I{P'm)-Yl{P"m') (i) 

The values of m and m' may be expressed as follows : 

CD-AJ CD-x 
^~ AC ~ V ' 



, CD-BG CD-x' 
^ ~ BC ~ I" ' 



...M.C.[.(^> .(-')]. 



. (2) 



If the loads advance a distance dx to the left, the value 
of M becomes 



M 



.. ^r.( VP'{x~Ax)-\ YP"{x' + Ax)'\] 



Art. 7.] MAXIMUM MOMENTS IN A BEAM. 7 1 

The change in the value of il/ is therefore 

For a maximum or minimum J7l/=o, hence 
IP' IP" 



V I' 



is> 



If IP represents IP' -{- IP" , then eq. (5) may take 
the form 

IP I 

xP=r'- • • . ^- . • • (6) 

Eq. (6) is the criterion desired; it represents an equa- 
tion whose conditions must be fulfihed for maximum 
moment. 



Art. 7. — Maximum Mcments in a Beam. 

It should be observed that the influence line for reac- 
tions, as found in Art. 3, is a ftmicular polygon for which 
the pole distance is the perpendicular distance between 
the pole L and the line I\IK. This polygon dift'ers from 
the ordinary funicular polygon, however, in that the various 
loads and distances are laid off in order exactly reverse 
to the usual procedure ; that is, the loads are laid off 
upwards beginning with Pi, and the distances a, b, etc., 
are laid off from right to left, beginning at the right with 
a. This brings the hea.d cf the moving load to the right, 
whereas in the usual procedures the head of the moving 
load is at the left. The significance of this construction 
should be carefully noted. 



72 INFLUENCE LINES. LLh. II. 

In finding the maximum moments in a beam, it will 
be necessary to use the ftinicular polygon as a moment 
polygon in the case of parallel forces and also to use the 
criterion deduced in the preceding article. 

In order that the condition of equilibrium expressed 
by that equation may hold, it will usually be found necessary 
to place a load directly at the section, since any portion 
of this load may be considered to be on either side of the 
section. This criterion is easily adapted to graphic con- 
struction. 

In Fig. 3 of Art. 3 draw LE equal to P\ vertically 
above L ; from E draw a horizontal line until it meets at F 
an ordinate dr£:wn vertically from R. Lay off EG equal 
to P2', from G draw a horizontal line GH, etc.; follow 
this same form of construction for the rem.ainder of the 
oads. The result will be the stepped diagram LEEGH . . . 
representing graphically the summation of all the loads, 
and also of all the distances from the head of the moving 
load to any point in the moving system. 

In the case in hand it is desired to find the position of 
the moving load causing the maximum bending moment 
at any section CD distant m from the left abutment. 
Lay off to scale on the edge of a strip of paper the length 
of the beam /, and from the right-hand end of / lay off m. 
It is important to notice that although the section is 
distant in from the left abutment, the distance m must 
be measured from the right-hand side on the funicular 
diagram, since this polygon is in a reverse position to the 
usual. Move the strip of paper until the load P2 or EG 
is over the section under consideration. If this position 
gives a maximum bending mom.ent, the criterion maust be 
satisfied. Erect vertical ordinates ai the ends of the beam 
and at the section until these ordinates intersect the stepped 
diagram at A', B', and /. Draw a line from A' to B' ; 
if the line A'B' passes through the step of the diagram 



Art. 7.] MAXIMUM MOMENTS IN A BEAM. 73 

representing the load situated at the section, the criterion 
will be fulfilled and this position of the loading will cause 
maximum bending. For, in the similar triangles RJA' 
and B"B'A', 

RJ RA' . IP' I' 

B''B''A'B"' ^^^^^^^ IP ~ V 

Should the line A'B' not pass through the step GF, the strip 
of paper must be moved until another load is brought to 
the section and the construction above described must be 
repeated. 

If, as in the present case, the criterion is satisfied, the 
bending moment can be immediately obtained. Erect 
vertical ordinates at the ends of the beam until they inter- 
sect the influence line at B'" and A'. Connect these points 
of intersection. As in Chap. I, Art. lo, the value of the 
vertical intercept between this line B'" A' and the influence 
line, when multiplied by the pole distance, will give the 
bending moment at any section. In this way the one 
diagram can be made to serve in finding both maximum 
shears and maximum bending moments in a beam. 

In actual practice it may be foiond advisable in con- 
structing the influence line to take as a pole distance not 
the entire length of the beam., but only a fractional part 
of it. This will cause the influence line to have a steeper 
inclination and will tend to make more accurate the measure- 
ment of all vertical heights, which evidently will be all 
increased in the same ratio. Care must be taken in such 
a case to give to all ordinates, whether for shear or bend- 
ing moment, their proper values. 



74 INFLUENCE LINES. [Ch. II. 



Art. 8. — Maximum Stresses in the Web Members of a Truss 
with Parallel and Horizontal Chords. 

The graphical operations of the preceding articles are 
sufficient to furnish all the maximum stresses in trusses 
with parallel and horizontal chords. In finding the max- 
imum web stresses use will be made of the following 
principle : The stress* in any web member of a truss having 
parallel chords is equal to the shear in the panel multiplied 
by the secant of the angle which the member makes with 
a vertical. If the shear is known, the stress can be found 
very simply by a graphical construction. Let the vertical 
side of a right-angled triangle represent the shear to scale, 
and let the hypothenuse make an angle with this line equal 
to the inclination of the web member with a vertical. Then' 
the hypothenuse will represent the stress in the member 
to scale. 

Therefore, in order to find the maximum stress in a 
web member, it will be necessary to find the maximum 
shear in the panel in which the member is situated, and 
this shear will be found graphically by means of an in- 
fluence line differing but slightly from the influence line 
for maximum shears in a beam. The change involved 
arises from the fact that the loading on a truss is applied 
at the panel points of the chords. It is clear that this 
application of the loads affects only the panel under con- 
sideration, for in the other panels the loads, whether con- 
sidered as they stand or as if concentrated at the panel 
points, cannot influence the shear in the panel under con- 
sideration. 

The shear in any panel is constant between panel 
points, and is equal to the reaction at the left end of the 
truss minus all the loads which may be situated in the 
panels to the left of the panel under consideration, and 



Art. 8.] MAXIMUM STRESSES IN IVEB MEMBERS. 75 

minus that portion of tlie loading situated in the panel 
itself which is transferred to its left end. It is evident 
that for maximum shear no load should pass the panel 
itself. There only remains to be considered the distance 
which the locomotive must advance on the panel in ques- 
tion. 

As an illustration, the maximum stresses in some of 
tJie web members of an 8-panel, 208-foot Pratt truss will 
be determined, using as the loading the locomotive con- 
centrations designated in Cooper's specifications as £40, 
and assumed to advance from right to left. 

Let AB, Fig. 5, represent the truss under considera- 
tion. Draw the base line KL equal in length to the total 
length of the truss. Beginning at L and with wheel i lay 
off on KL and towards the left, to scale, the distances 
between the various wheel loads, and erect vertical ordi- 
nates at the points thus found. It should be noted that 
the uniform load is treated as a series of concentrated 
loads, each concentration representing ten feet of uniform 
load. At K erect the vertical KM, and lay off on KM 
upwards, to scale, the amounts of the loads, beginning at 
K with wheel i. Connect the points thus found with L, 
which becomes the pole of the funicular polygon, and 
by the aid of these lines draw the funicular polygon 
LD" . . . N . As already demonstrated, this is an influence 
line for reactions and for shears for a beam with the span 
KL. 

In order to use this line in the present case, it will be' 
necessary to find the reactions at the left-hand end of 
any panel when loads are found within such panel. For 
this purpose, a similar influence line must be drawn, using 
in this case, however, the panel length as the length of 
span. This second influence line is shown in the lower 
left-hand comer of the figure as RE; in this case R is the 
pole. 



76 INFLUENCE LINES. [Ch. IT. 

In order to find the maximum stress in any member, 
such as 6-4, for example, it will be necessary to find the 
maximum shear in the panel 3-4: the two influence lines 
already constructed are then to be combined for this pur- 
pose. From panel point 4 let fall a vertical ordinate upon 
the line LD" . . . N, cutting it at G. GH represents the 
shear at the panel point 4, and also in the panel 3-4 when 
wheel load i is at th,is point. If the load advances so that 
wheel load 2 is at this panel point, the shear cannot be 
measured directly from the original influence line LD" . . . N 
Avith its value diminished by wheel load i ; the shear is now 
dininished only by that portion of wheel load i which 
is carried to panel pomt 3. The value of this portion is 
known from the line RE, and its position relative to the 
panel point 4 is also known. Therefore this value is 
drawn at the proper point as a negative ordinate from the 
line LD" . . .N. 

The same construction is followed for other positions 
of the loading, in which other drivers are at panel point 4. 
The final influence line is GSU, and ordinates between the 
base line KL and GSU will then show the shear in the 
panel, as the loading advances toward the left. By in- 
spection it is then seen that the maximum shear is TS; 
by employing the construction noted in the beginning of 
this article, the line SU\ drawn parallel to b-4, will represent 
to scale the stress in b-4. The stress in the vertical 6-3 
is found at the same time and is equal to the maximum 
shear in the panel or TS. 

A similar method of procedure is followed in the case 
of all the other web members, including as such also the 
end post. The graphical operations may be performed 
very expeditiously by treating the same load at one time 
in all the panels. By this means the counter-stresses in 
the members are also found at once, for the stresses in 
those members situated to the right of the centre of the 



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Art. f).] MAXIMUM STRESSES IN CHORD MEMBERS. 77 

truss indicate the maximum compression which can exist 
in the corresponding members to the left of the centre; 
For convenience the stresses in these members have been 
found as if they sloped downward to the right instead 
of downward to the left. 

The dead -load stresses can be found in the usual manner 
by a force polygon, or by representing the shears in the 
truss by a stepped diagram, and using the construction 
employed above. By comparison with the live -load 
stresses, the necessity of counter -braces is at once deter- 
mined, and also the values of their stresses. 

Art. 9. — Maximum Stresses in the Chord Members of a Truss 
with Parallel Chords. 

In order to find the maximum stresses in the chord 
members of a truss with parallel chords, the principle of 
sections will be employed. The method used, however, 
can only be applied, as far as the present analysis is con- 
cerned, to those trusses in which the centre of moments 
is found in a vertical line drawn through a panel point of 
the loaded chord. The truss and loading represented in 
Fig. 4 will again be employed and the maximum stress in 
the upper chord member b-c will be determined. Panel 
point 4 is the centre of moments for this case; therefore 
the maximum moment at the point 4 must be found. As 
already noted, the funicular polygon LD" . . . N is in a 
reversed position to the loading as it advances on the 
truss from right to left. Let the line BA below the line 
KL represent the truss laid off to scale on a strip of paper, 
the point 4 representing the centre of moments. By use 
of the stepped diagram, explained in Art. 7, it is found 
that wheel 9 at the section gives one position for maximum 
bending. By erecting ordinates AA' and BB" and drawing 
the line A'B", the bending moment CD" is graphically 



78 



INFLUENCE LINES 



[Ch. II. 



obtained, but it must be multiplied by the pole distance 
in order to be expressed in proper units. The maximum 
live-load stress in b-c is the bending moment thus found 
divided by the depth of truss. The dead -load stress is 
found either by a force polygon or by means of a moment 
curve. 



Art. 10. — Influence Lines between Adjacent Panel Points. 

Beams and girders as a rule carry the loads imposed 
upon them at any point in their spans. In the usual types 

of trusses, however, and sometimes 
even in the case of long girders, 
the load is brought upon the 
structure at fixed panel points 
only. Due to this redistribution 
of the load at panel points, it be- 
comes necessary to investigate the 
change occurring in influence lines 
which may have been drawn under 
the assumption that a load acts 
where it appears to be placed. 
Let Fig. 6 represent two adjacent 
panel points designated as points n—i and n, and let the 
load P be distant x from n. The effect of the load P on 
the members of the truss will be that due to its distribution 
to the adjacent panel points. In this case the equivalent 
effect is that due to forces equal in amount but opposite 
in direction to the reactions of a load F on a span of panel 
length /. If these forces are represented by Rn-i and R, 
there will be found 

-1 = ^-; (i) 




Fig. 6. 



i?«-i=-^ 



R = 



P(l-x) 
I 



(2) 



Art. II.] MOMENT INFLUENCE LINES FOR ^NY TRUSS. 79 

Let the ordinates of the influence hne beneath the 
panel joints and beneath the load be designated respect- 
ively by y„-i, >'„, and y. Since the product of the load P 
by its influence ordinate must be equal to the sum cf the 
products of its parts, each m.ultiplied by its coiTesponding 
coordinate, there will be obtained 

Py^R„.i-y,,^i + R„-y„ (3) 

By the substitution of the values of eq. (i) and eq. (2) 
in eq. (3) there will be found 

X I — X 

y-=jyn-i + -j-yn " . (4) 

In this expression x is the variable quantity, and since 
it appears only in the first degree, the line which eq. (4) 
represents must be a straight line. The following rule 
may therefore be deduced: An influence line for a single 
moving load showing the variation in any function for 
any part of a truss is a straight line between adjacent 
panel points. 



Art. II. — Moment Influence Lines for Any Truss. 

If the centre of moments lies on a vertical which in- 
tersects the loaded chord line within the limits of a panel, 
the moment influence line treated in Art. 5 must be 
modified in consequence of the redistribution of the loads 
within the panel to the panel ends. 

In finding the stress in the chord member BD, Fig. 7, 
where the centre of moments C falls within the panel BD, 
the influence line drawn without reference to the existence 
of the panel is only correct for the portions of the truss 
AB and DE, since in finding the moment at C the loads 



8o 



INFLUENCE LINES. 



[Ch. II. 



on these portions of the truss are not redistributed. It 
has been shown, however, (Art. lo) that all influence lines 
for single loads between panel points are straight lines; 




Fig. 7. 

therefore that portion of the influence line which lies within 
the panel is then constructed by connecting / and F, the 
points of intersection of the influence line already drawn 
with the verticals erected at the panel ends. The in- 
fluence line is then EFJA. 

In order to find the position of a series of concentrated 
loads causing the maxiraum moment at the section C, it 
will manifestly be incorrect to use the criterion found in 
Art. 6, but the proper criterion is easily deduced. 

The following notation (Fig. 7) will be employed: 

/' and /" =the distances from the left and right abutments 

to the centre of moments respectively; 
I =the length of span; 
p =the length of panel; 

q =the distance from the centre of moments to the left end 
of the panel; 



Art. II ] MOMENT INFLUENCE LINES FOR ANY TRUSS. 8i 

P and P' =any loads on the sections AB and DE respect- 
ively ; 

P" = any load within the panel BD ; 

:j;, :jf', and y =the respective distances of P, P' , and P" 
from the right ends of the various sections ; 

m, m', and w=the respective influence ordinates for P, P' , 
and P". 

Then, as usual, the momer.t M for any position of the 
loads becomes 



M = I{P-m)-Vl{P"-n)^-I{P'-m'). . . (i) 

If the ordinate n is divided into two parts n' and n" 
by the line BF, then the moment Al for any position of 
the loads, using a notation similar to that before, becomes 

M = I(P-m) + I{P"-n') + I{P" -n") + I{P'-m'). (2) 

Eq. (i) may take the following form by substituting 
for the influence ordinates equivalent values fotmd from 
the similar trangles of Fig. 7: 

If the loads advance a distance Ax to the right, the 
change in the values of the moment becomes AM and is 
equal to 



82 INFLUENCE LINES. [Ch. IL 

= l\p—I^Ax'\-^l\^{ql-l'p)Ax'\-l\p'jAx'\. 
For a maximum or minimum AM = o ; noting that 

4-t]-[-I]-[4]. 

there is found 

i'P-i-[(P + P' + P")y] + ^[p"|]=o. . . (4) 

If IW represent all the loading on the span, then 
eq. (4) will take the form 

IP + pP"JjIW (5) 

This criterion is used in the same manner as the crite- 
rion deduced for finding maximum mom_ents in a simple 
beam, but it differs from that in one respect, viz., the 
left-hand merpber of eq. (4) does not include the sum of 
all the weights in the panel, but only a definite portion. It 



Art. 12.] yARIATION Oh' MOMENT IVITHIN A PANEL. 03 

will usually be found that the conditions of the equation 
are fulfilled by placing a wheel load at the left end of the 
panel. 

Art. 12. — Variation of Moment within a Panel for a Fixed 
Position of the Loading. 

Before proceeding with an example showing the appli- 
action of the criterion just deduced, it is advisable to show 
that if a series of concentrated loads is fixed in position 
on a truss, the variation of the bending moment between 
panel points may always be represented by a straight line. 
Let it be required to find the variation of bending moment 
in the panel p of the truss shown in Fig. 8, the position of 




Fig. 8 



the wheel loads being fixed. Let R be the reaction at A, 
the left end of the truss, for all the loads shown on the 
truss, and R\ the reaction at the left end, C, of the panel 
CD, for the loads within the panel; the other notation 
to be employed is shown in the figure. Then 

i?=Pi[(a + 6 + c4-... + w)+P2(& + c + (i+... + w) ... 

R\ =[-P3(c + w') +P4(nO ...]-, n' being the distance between 

P 
P4 and the right end D of the panel. 

If M represents the moment at any point in the panel 
p distant x from A , 

M = R-%-Px{oi-ni)-P-i,{x-a-m) . . . -R^ix-V). (i) 



84 INFLUENCE LINES. [Ch, II. 

Eq. (i) shows that between panel points ilf . varies 
directly as the first power of x, and that the variation may 
consequently be represented by a straight line. There- 
fore to find the moment at any section within a panel, for 
a fixed position of the loading, it suffices to know the 
moments at the panel ends; by connecting the ends of 
the ordinates representing these end moments by a straight 
line, any intermediate moment is at once obtained. This 
method is easily applicable with a funicular polygon, since 
the latter is a moment polygon for parallel loads. 

Art. 13. — Problem in Finding the Maximum Stress in the Loaded 
Chord Member of a Truss with Web Members all Inclined. 

An application of the principles of the preceding articles 
will be made in finding the maximum stress in the loaded 
chord member 3-4 of the 6-panel deck-truss shown in 
Fig. 9. The following data are given: 

Panel lengths, all equal 29 feet 

Length of truss 174 

Depth of truss at panel point a. . . . 19 

Depth of tniss at panel point b 24 

Depth of truss at panel point c 25 " 

Locomotive loading: E 40. Cooper's Specifications. 

The reaction influence line LMNPQ . . . and the stepped 
diagram LABCDE . . . are drawn in the usual manner 
The truss itself is redrawn on a separate strip of paper, 
but in the reverse position to that of the upper diagram 
the reason for which has already been explained (Art. 7) 
It is shown in dotted lines in the lower part of Fig. 9 
This strip of paper must then be placed in such a position 
that the criterion of Art. 11, 



Art. 13.1 THE MAXIMUM STRESS IN LOADED CHORD STRESS. 85 

is satisfied with respect to the member 3-4. 

In this case, q is the distance 3 to c' ; p is the distance 
3 to 4; /' is the distance 1 to c' ; I is the distance i to 7. 

Wheel 8 will rest at panel point 3 if the truss be placed 
in the position shown in the lower part of the figure. Verti- 




cals erected at the ends of the truss and at the panel points 
3 and 4 will intersect the stepped diagram at the points F 
E, A, and C respectively. 

1 2W is then represented graphically by EE' , and ^B 
either by A' A or A'B, since wheel 8 may be taken to act 



86 INFLUENCE LINES. L<-»- '^'^■ 

either to the right or left of the panel point. For a similar 
reason, IP" may be represented correspondingly either 
by A"C or B'C. From similar triangles, 

GG' : EE':: FG' : FE' ; 
that is, 

GG'=jIW. 

According to the criterion, GG' must equal either 

A'A^^A"C 
P 

or 

A'B + iB'C. 
p 

From similar triangles it is seen that 

^A"C=A"'G'", and that ^B'C=B"G". 
P P 

Therefore, by substituting these values in the criterion, 
the value of GG' should equal either G'G'" or G'G" . In 
the present case G'G lies between these two values; it is 
evident that if wheel 8 be divided into proper parts at 
the panel point, the conditions of the criterion will be 
exactly fulfilled. In practice, therefore, it is only necessary 
to determine whether the point G lies between the lines 
BC and AC; if it does, as in the present case, this position 
of the loading will furnish one maximum value for the 
stress in 3-4. 

It will usually be found that more than one position of 
the loading will fulfil the required conditions; in such a 
case the actual values of the stresses must be determined 
and the absolute maximum taken. 



Art. 14.J STRESSES IN THREE NON-CONCURRENT MEMBERS. 87 

The value of the moment at C is foiind by means of the 

method of Art. 12. The closing line FQ of the funicular 

polygon is drawn and the points M and N, corresponding 

to the panel points 3 and 4, are connected by a straight 

line; the intercept XY when multiplied by the proper 

pole distance will give the value of the moment; in this 

case 8,734,000 foot-pounds. The stress in 3-4 is this mo- 

-,■ • 1 1 1 1 1 /I • 8734000 
ment divided by the lever-arm cc ; that is, — ^^ = 

349,000 pounds compression. The stresses in the other 
members of the loaded chord will then be found in a pre- 
cisely similar manner. 

Art. 14. — Determination of Stresses in Three Non-concurrent 
Members of a Truss. 

A frequent problem * in statics is the determination of 
the stresses in three truss members not meeting in a 
point. It is assumed that the external forces are fully 
known in regard to the magnitude, direction, and point of 
application. In the following treatment these external 
forces may therefore be replaced by their completely 
known resultant. Let the section mn, Fig. 10, cut the 




Fig. 10. 

three members U, L, and D of the truss, and let R repre- 
sent the position, direction, and magnitude of the re- 
sultant of the external forces situated at the left of the 

* See Chap. I., Art. 12, for a solution of the same problem, but stated in more 
general terms. 



■88 INFLUENCE LINES. [Ch. II. 

section and which holds in equiHbriiom the stresses in the 
members in question. 

Continue L till it intersects R at a. Then connect h, 
which is the intersection of U and D, and a by the line C, 
and which, for the present, may be assumed the line of 
action of a supplementary force acting through a and h. 
The force polygon stn may then be drawn by using the 
three concurrent forces R, C, and L, since only the amounts 
of C and L are unknown. This operation can be repeated 
with the forces C, U, and D, assuming in this instance C 
to be completely known.- The resulting triangle is suv. 

If C in the second case be given a direction opposite 
to that in the first case, it is found that the supplementary 
force C is annulled; in other words, the resultant of two 
of the forces balances the resultant of the other two. Thus 
the forces R, L, D, and U form a closed polygon, and the 
forces which are represented by the sides of this polygon 
are in equilibrium. The stresses D, L, and U are then 
completely determined. 

Art. 15. — Stresses in the "Web Members of any Simply Supported 

Truss. 

In order to determine the variation of the stress in the 
web member or diagonal D of any truss such as shown in 
Fig. 1 1, in which the moving load traverses the lower chord, 
let a load P move over the truss from the right abutment 
to panel point 3. By the- method of sections and moments 
it can be seen that the stress in D in this case is influenced 
simply by the reaction R at the left abutment ; it is clear, 
therefore, that with the load P between panel points 3 
and 5 the stress in D will always be of the same sign. 
In the same way for a load P between the left abutment 
and panel point 2 the stress in D is influenced only by the 
reaction R' . This stress will always be of an opposite 



Art. 15.] STRESSES IN IVEB MEMBERS OF SIMPLE TRUSSES. 



89 



kind to that with the load between the points 3 and 5. 
It should, however, be noted that these statements do not 
apply to trusses of that kind in which the two chord mem- 
bers which are cut by the section meet within the limits 
of the truss.* In the f.dlowing treatment, therefore, tinisses 
of this character are excluded. 

It must be evident that some point in the panel 2-3 
(Fig. 11) must be a critical point for deciding how far 




Fig. II. 



a load may advance into the panel before the stress in the 
diagonal of that panel is reversed. This critical point 
will be found where the load if applied will cause no stress 
at all in the member. The following construction will 
determine its location. 

Continue the unloaded chord member a-h until it meets 
the lines of action of the reactions at g ajad h\ then draw 
the lines gk and hk through the panel points 2 and 3 to 
the intersection k. A vertical line through k will deter- 
■mine the desired critical point, as will be shown. Assume 
any load P placed vertically over k. Its effect on the 
stresses in the members of the truss is the same as if it 

* See Art. 22 of this Chapter. 



9° INFLUENCE LINES [Ch. II. 

were replaced by its components P2 and P3 acting at 
panel points 2 and 3, the amounts of which are found by 
means of the triangle 2^3, a funicular polygon for this 
loading on a span equal to the panel length. The force 
polygon corresponding to this funicular polygon is shown 
in Fig. 1 1 , and represents P2 and P3 to scale. In a similar 
manner ^23/2 is a funicular polygon for the forces R, R', 
P2, and P3, and the values of the reactions R and R' of the 
main truss are found on the same force polygon, Fig. 11, 
by drawing a line through the pole of the force polygon 
parallel to ab. The forces whose moment is to be taken 
and which, cause stress in D are therefore R and P2. 
Their resultant, whatever be its value, acts at the inter- 
section of the strings ab and 23 of the funicular polygon. 
But this point of intersection is also the intersection of 
the two chord members U and L, and is the centre of 
moments for finding the stress in D. The resulting moraent 
of R and P2, therefore, is zero and the diagonal sustains 
no stress for this position of the loading. Therefore loads 
advancing from R' to the. critical point K will cause stress 
of one kind, and loads from R to K stress of the opposite 
kind. Trusses with the loading on the upper chord are 
treated in a precisely similar manner, the construction to 
be followed in any case being simply this: Continue the 
unloaded chord member cut by the section till it intersects 
the lines of action of the reactions; then draw lines from 
these points through the ends of the loaded chord member 
which is cut by the section, and the intersection of these 
lines will give the desired point. 

If the moving load is a uniform load, its position for 
the maximum stresses of opposite kinds is immediately 
found; the loading extends alternately from either point 
of support to the critical section. The value of the stress 
itself may be found quickly in a manner which, although 
approximate, is sufficiently exact to cover the usual cases 



Art. 10 J INFLUENCE LINE FOR ANY IVEB MEMBER. 91 

occurring in practice. The stresses so found will have a 
value slightly greater than those which actually exist. 
Assuming the load to advance from the right abutment, 
the approximation consists simply in neglecting that por- 
tion of the loading acting downward at panel point 2, 
and considering the stress in the diagonal to be caused only 
by the reaction R. The value of R may be found very 
simply either analytically or graphically. By means of 
the construction explained in Art. 14 the stress in the 
member may then be found as follows : 

The reaction R acting at i holds in equilibrium the 
three unknown forces U, D, and L. Continuing U and L 
to intersect at m, and letting D intersect R at n, the force 
acting along the line nui is found to balance these two 
pairs of concurrent forces. In order to obtain the value 
of D, lay off upwards frorn n the value of R equal to np. 
Through p draw a line parallel to D to intersect nin at q; 
pq will then be the stress in D, and will be completely 
determined in regard to amount and direction. In order 
to find the greatest counter- stress, the load must cover that 
portion of the truss not loaded before; otherwise the 
operation is precisely similar to that just described. 

Art. 16. — Influence Line for Stress in any Web Member of a 
Simply Supported Truss.* 

Method I. 

As a unit load passes over the bridge from right to left, 
an influence line for the stress in any diagonal, such as AC 
of the truss shown in Fig. 12, may easily be drawn. In 
this case the load is carried at the panel points of the lower 
chord, but the methods to be used are also applicable to 
trusses in which the load is carried by the upper chord. It 

* That type of truss, however, in which the chord members intersect within 
the limits of the span is again excluded (see Art. 15). 



92 



INFLUENCE LINES. 



[Ch. II. 



is clear that all loads between points R' and B will cause 
stresses in the diagonal of the same sign, and their aggre- 
gate magnitude will vary directly as the left-hand reaction. 
The variation of this reaction can be indicated by an influ- 
ence line for reactions, and is a straight line for a unit load. 
Therefore the influence line for stress in the diagonal AC 
for a single load may be drawn upon the reference line 




Fig. 12. 



MN, as a straight line NL between R' and B; its slope 
must be fixed by locating two of its points. 

Similar reasoning holds for the distance RC. Hence 
the influence line for this portion of the truss is also a 
straight line, but since the stress is of an opposite kind 
from that when the load is between R' and B, the line 
MK is drawn below the reference line. There still remains 
to be determined the influence line when the load is in the 
panel itself. 

It has already been shown (Art. 15) that there is a 
point D in the panel where a load, if placed, produces no 
stress in the diagonal of that panel ; and it has been shown 
(Art. 10) that all influence lines are straight lines between 



Airr. i6.] INFLUENCE LINE FOR ANY IVEB MEMBER. 93 

panel points. Therefore the influence line in the panel 
itself is found by drawing any straight line through D' 
found vertically below D. The line KD'L intersects verticals 
drawn through the panel points K and B. The influence 
line is then completed by connecting the points K and L 
with the ends of the reference line M and A'', since it has 
just been shown that these portions are straight lines, and 
two points upon each of these lines have now been found. 
Ordinates above the reference line indicate stress of one 
kind, and ordinates below, stress of the opposite kind. 
The scale with which to measure the influence ordinates 
remains to be determined.; but it is clear that if the stress 
is known for one position of the load, such as B, the scale 
is immediately fixed. 

The influence line for a unit load having thus been found, 
the position of a series of concentrations causing the greatest 
stress can be found precisely as in Art. 2. The loads 
having been placed in a trial position, the sum of the 
products of the ordinates of the influence line by the amount 
of the load placed over any ordinate may be found and 
compared to another trial position. The greatest sum 
will give the position of the loading for maximum stress 
and, by means of the proper scale, the stress itself. It is 
possible, however, by the aid of the influence line just 
developed, to deduce a criterion which will at once enable 
the position of the loads causing the maximum stress to 
be determined. (Art. 17.) 

The ease with which uniform loads may be treated is at 
once evident from Fig. 1 2 ; for maximum stress it is only 
necessary to cover the portion ND' of the truss; knowing 
the scale of the diagram, the stress in AC is the area D'LN 
= ^D'NxGL. These distances are easily measured from a 
carefully executed diagram. 

The scale of a stress influence diagram may be found as 
follows: Let there be placed at any panel point of the 



94 INFLUENCE LINES [Ch. II. 

loaded chord a unit load, and let there be drawn for this 
one load an ordinary stress diagram for the structure 
furnishing the stress for each member for the one load 
employed. These stresses provide at once a means of 
determining the scale of all influence-line diagrams, for 
they furnish for each influence line the value of the 
ordinate below the point of application of the load. 

The use of influence areas becomes immediately avail- 
able in actual design work, if an equivalent uniform 
load could be found to replace locomotive concentrations. 
Such an equivalent load, however, would vary not only 
with the span lengths, but also with the purpose for which 
it is to be used, that is, for moments or shears. It is 
becoming more the custom, however, in bridge-design 
offices to obtain an equivalent uniform loading for every 
span length, and for all conditions. Once obtained and 
tabulated, they may be quickly applied in all influence- 
line work. 

Method II. 

The variation of the stress of a web member, such as 
CG, in the truss shown in Fig. 1 3 , may under certain con- 
ditions be more conveniently represented by the aid of the 
following analysis: 

/i =that portion of the span length between the left abut- 
ment and the left end of the panel cut by the section ; 

/2=that portion of the span length between the right 
abutment and the right end of the panel cut by the 
section ; 

p = the panel length ; 

/=the length of truss; 
w=the distance between the centre of moments and the 
left abutment ; 

X =the distance of any load from K, the centre of moments ; 



Art. i6.J INFLUENCE LINE FOR ANY IVEB MEMBER. 



95 



Ra and Rb = the left and right reactions respectively ; 
d =the lever-arm of CG about E (not shown) ; 
S^and 55=the stress in the member for a single load on 
each span section /i and I2 respectively. 




Then, for a unit load on /i, the general expression for 
Sa becomes 



Sa = 



Rsil + m) 



(I) 



But Rb ■■ 



x—vn 



I ' 



Sa- 



(x — m)(l+'m) 
Td • 



(2) 



Similarly, for a unit load on the span I2, the general 
value of Sb becomes 



5b = 



Ra-wi 
~d~' 



96 






INFLUENCE LINES. 


But Ra 


l + m- 
l 


-X _ 


. ^ (l + m — x)m 



[Ch. II. 



(3) 



The variations expressed by eqs. (2) and (3) may in the 
usual manner be represented by straight lines. But when 

X = the values of Sa and Sb each become equal to fj — ; 

this means that the straight lines intersect or cross at the 
origin of ordinates (also origin of moments) where x = o. 
Following this reasoning the influence line fjr stress in 
CG may at once be drawn; through any point, as U, in 
the vertical line through E draw the lines UV and UT 
to the ends of the span. AVhen x=m, 5.4=0; and when 
x = l-\-m, Sb = o. Hence eqs. (2) and (3) represent the 
lines UYT and UVW. Dropping the verticals CW and 
DY and connecting W with 1', the influence line VWXYT 
at once results, for within the panel CD this line is straight. 

Whenever the centre of moments falls within convenient 
limits on the drawing. Method II is preferable to Method I, 
as' it is general and holds whether the centre of moments 
falls within or without the limits of the span. 

The scale with which to measure the influence line 
is easily obtained, for, after having obtained the stress in 
CG for a load placed at any point, such as D, the scale 
for the remaining ordinates is at once determined. 

It should be noted that this analysis is also directly 
applicable for drawing the influence lines for the stresses 
in -the chord members. 



Art. 17.] POSITION OF LOADING FOR MAXIMUM l-^EB STRESS. 97 

Art. 17. — Criterion to Determine Position of Loading for 
Maximum V/cb Stress. 

The web member CA of the truss shown in Fig. 12 will 
be chosen, and it will be supposed that the loads advance 
upon the bridge, along the lower chord, from right to left. 
It is at once seen, by inspection of the influence line, that 
in general no load must pass the panel in question, since 
loads in that portion of the truss to the left of the panel 
will always cause negative stress ; the exact distance which 
the loading must advance into the panel is then the quantity 
which must be determined. Let / be the distance from 
the end of the span R' to the point D at which a load causes 
no stress in CA ; /' the distance from the right end of the 
panel to the same point D; x the distance of any load P 
from the end of the span; x' the distance of any load P 
from the right end B of the panel; ni the general value 
of the influence ordinate between R' and B corresponding 
to a unit load; {m — in') the value of the similar quantity 
between B and C; a the value of in' at the point D; IP 
all the weights on the bridge; IP' the weights on the 
•panel BC; and S the stress in the member AC. Evidently 

S = IPm-IP'm' (i) 

ax 



From similar triangles there is found that m = -r- and 

ax' 
fn' ^—jf . Substituting these values, eq. (i) becomes 

S=a\jIPx-j,IPx'\ -. (2) 

If the train advances to the left by an amount equal 
to Ax, the stress becomes 

S'^a\jIP{x^-Ax)-jIP{x' + Ax)\, 



98' INFLUENCE LINES. [Ch. 11. 

and the change in the stress is therefore 



AS=S'-S^a\jIPJx-j,IP'Ax\, . 



(3) 



assuming that no new loads advance upon the truss, and 
that no new loads enter the panel. 

For a maximum or minimum i5 = o ; hence for this 
condition 

I- IP ^ • U) 

This equation is a perfectly general equation of condi- 
tion for finding a maximum stress, and is made applicable 
for finding the greatest counter stress by substituting for 
/ and /' the proper quantities measured from the left ends 
of the truss and panel. 

In order that eq. (4) may hold, it will usually be found 
necessary to place a weight at the panel point B, and to 
consider only so much of it in the panel CB as may be 
necessary to fulfil the conditions of the equation. 

This criterion may be used graphically precisely as in 
the case of moments in Art. 7. Lay off the weights Pi, 
P2, etc., on 'a vertical line passing through A^", Fig. 12, 
beginning at the bottom with Pi, and on a strip of paper 
lay off the distances between the loads, beginning at the 
right with Pi. Let NE represent the total load IP on 
the bridge, P2 being at the panel point B. Connect D' 
with ii by a straight line cutting a vertical erected at B 
or P2at F. If a horizontal line drawn through F inter- 
sects the load P2 on the line NF the equation of condition 
is satisfied, and this will be one position of the loading, 
causing maximum stress. For, from similar triangles, 

lyG^NH V_ IP^ 

DN'NE °^ l^IP- 



Art. 17.] POSITION OF LOADING f-OR MAXIMUM IVEB STRESS. 99 

Application of the Criterion. 

An application of this method to find the maximum and 
minimum stresses in the diagonal 54 of the truss shown in 
Fig. 14 will now be made. The following are the required 
data: 

Span = 264 feet; number of panels = 8; panel length 
= 2,:^, feet; depth of truss at centre = 42 feet; length of 
53=39 feet; length of 5i=3i feet; loading = A" 40 of 
Cooper's specifications. 

Following the methods of Art. 15, it will be found that 
D is the point where a load, if placed, will cause no stress 
in 54, and this point is projected vertically downward 
as D' upon the base line MN . It will be supposed that 
the loading has advanced upon the span from the right, and 
to such a point that wheel load 3 is at the panel point B. 
The number of concentrations on the span, as well as the 
position of each load, is easily found by laying off to scale 
on a strip of paper the distances between the loads and 
moving this strip lontil w^heel 3 is at the point G. The 
line MN shows this particular position of the loads. It 
should be noted that 10 feet of uniform load are treated 
as one concentrated load, the effect of the latter being 
to act at the centre of the lo-foot section. The amounts 
of the wheel loads are then laid off upward in regular 
order on the vertical erected at the right end of the base 
line, extending from A'^ to E. The points D' and E are 
then connected by a straight line and a horizontal FH 
drawn through F, the intersection of D'E with a vertical 
erected at G. Then, since FH intersects the load line in 
the load 3, the conditions of eq. (4) are fulfilled, and this 
position of the loading causes a maximum stress. 

In order to determine the actual stress in 54, it will then 
be necessary to find both the reaction at the left end of the 
span and the reaction at the left end of the panel; these 
may be found, both for this position and other positions 

UOFC. 



lOO 



INFLUENCE LINES. 



[Ch. IL 




Fig. 14. 



Art. 17.] POSITION Oh' LOADING FOR MAXIMUM IVEB STRESS. 101 

which the loading may occupy, from a reaction influence 
line NSTU for the main span and a line VW, shown in the 
left-hand comer, for the panel length. These forces XY 
and X'Y' are the only external forces acting at the left 
of the section passing through the members ^354 and L3; 
therefore their resultant holds in equilibrium the stresses 
in these three members and these stresses may be found 
by means of the method of Art. 14. 

At this point, however, an approximation involving 
but small error and that on the side of safety will be in- 
troduced and consists simply in neglecting the panel re- 
action X'Y'; that is, instead of taking the resultant of 
these two external forces, which acts a little to the left of 
the point R, only the reaction A"l" at R is taken. The 
point A, which is the intersection of the lines of application 
Us and 54 is then connected with the point R, which is the 
intersection of the lines of application of L3 and XY, and 
the force diagram PRLK drawn, the line RL representing 
to a much reduced scale the ordinate XY found from the 
reaction influence line below the wheel load i. By means 
of the proper scale, the maximum stress in 54 is thus found 
to be a tension of 157,000 potmds. 

In order to find the minimum stress, the maximum stress 
in the member 54' situated in a corresponding position on 
the other side of the centre of the truss will be fotind. 
By means of an exactly similar construction to that em- 
ployed above, it is found that wheel load 2 at panel point 
C will cause a maximum stress. The reaction is found 
to be X" Y" and the stress by means of the polygon R'P'K'U 
at the right end of the stress is found to be a compression 
of 52,000 pounds. The length R'L' is equal to X"Y" . In 
order to avoid confusion between main and coiuater stresses, 
the diagrams for main stresses should always be drawn at 
the left end, and those for the counter stresses at the right 
end of the truss. 



.I02 INFLUENCE LINES. [Ch. II. 

As already noted, the values of the stresses found in this 
rnanner are not quite exact; in order to make them ab- 
solutely so, the effect of the small forces acting at the left 
■end of any panel in question must be included and may 
be treated as if existing independently of the main reaction; 
•the quantity thus found separately must be added alge- 
braically to the stress previously found to make the latter 
exact. For instance, in the case of Si the quantity X'Y' 
is equal to 6500 pounds acting downward at C\ passing 
a section through t/3, 54, and L3 it is found that, due to 
X'Y', Si sustains a compression of 7000 pounds; the exact 
.stress in 54 is therefore 157,000 — 7000 = 150,000 pounds 
; tension. The percentage of error involved in the approxi- 
mation is never very large; the judgment of the designer 
must determine whether to apply the exact procedure or 
whether to make the approximation. Stresses in the 
•other web members may then be found by similar methods 
pf procedure. 

There still remain to be determined the stresses in the 

chords, which are found precisely as in the case of trusses 

■with parallel chords by finding the maximum bending 

moments at the various panel points and dividing those 

-moments by the proper lever-arms for the various chord 

.members. For instance, to find the maximum stress in 

L3 the maximum bending moment for the point A is 

(divided by the distance AC, and the quotient is the stress 

,:in L3. In this way all the live-load stresses maybe found; 

!the dead-load stresses, as usual, are to be found by means 

of the ordinary force polygon. 

; ,; The preceding work, involving the use of wheel-load 

.concentrations, requires that for every truss a separate 

reaction influence line NSTXU must be drawn. This 

::)-.equires careful draughtsmanship, and could be avoided 

jjf iequivalent uniform loads could be fixed upon, for then 



Art. i8. 



TRUSSES IVITH SUBDll/.DED PANELS. 



103 



the methods of Art. 16 would be simply applied, and 
influence areas would replace influence lines. 

Art. 18. — Trusses with Subdivided Panels. 

Trusses of long spans frequently have panels subdivided 
in the manner shown in Fig. 15. In the case there shown 




Fig. 15. 

the panel L3L5 is divided by the tie-rod M4L4, which is 
hung from the point ilf 4 and which distributes the weight 
it carries at its lower end to the main panel points of the 
truss by means of one of several subsidiary trussed panels. 
In this way, long stringers and consequently heavy floor 
systems are avoided. The stresses in the truss remain 
perfectly determinate, but the criteria previously deduced 
either need sorne additional explanation or require some 
modification to be applicable to this form of truss. It should 
be noted that U^Lq is the main web member of the panel 
L3L5 and that f/sLs is the counter member, but that por- 
tions of each of these web members are also included in 
the subordinate framing. The members MzM^, U4M4, 
etc., shown in dotted lines are not true parts of the truss; 
their functions are simply to support compression members 
at intermediate points, and thus by reducing their effective 
lengths permit the use of higher intensities of stress. 

In the treatment which follows it will not be necessary 
to discuss both the main and counter web members, since 



I04 



INFLUENCE LINES. 



[Ch. II. 



the same methods are applicable to both ; it is only neces- 
sary to remember that main members have their maximum 
stresses when the loading covers the longer portion of the 
truss and counter members when the loading covers the 
shorter portion of a truss. 

Considering, then, only the panel L3L5, Figs. 16 to 19, 
the dotted lines illustrate the various ways in which the 





Fig. 17. 

subordinate bracing may act to transfer the load at the 
point il/4 to the panel points L3, L5, ^3, and U5. In 
Fig. 16 the trussing L3M4L5 distributes the load to L3 and 
L5; in Fig. 17 the trussing UsMiU^ distributes the load 
to U3 and U5; in the former case the subordinate inclined 
bracing (shown in dotted lines) is in compression; in the 
latter, in tension. In Fig. 18, the trussing UzM^Lz dis- 




flG. 18 



tributes the load to f/3 and L3 and similarly in Fig. 19, 
U^MiL^ distributes the load to U i, and L5; in these two 
cases the subordinate bracing is partly in tension and partly 
in compression. The tie-rod sustains tension in all cases. 



Art. 19.] TRUSSES y/lTH SUBDIVIDED PANELS. 105 



Art. 19. — Maximum Web Stresses in Trusses with Subdivided 

Panels. 

It is at once seen that the bracing shown in dotted 
lines in Figs. 16 and 17 transforms the short stringers 
L3L4 and L4L5 into one trussed stringer, and that the loads 
on the panel length L3L5 are distributed only to the points 
Lz and L5. The criterion of Art. 17 may then be imme- 
diately applied as it stands, taking L3L5 as the panel 
length. The maximum stress in U^Lr, will then be found 
as the tension resulting from the position of the loading 
determined - by this criterion; but in Fig. 16 this stress 
must be decreased for Af 4L5 by the amount of compression 
which this portion -of UzL^ carries as a member of the 
trussed stringer. Similarly in Fig. 17 the stress in U^Mi 
must be increased by the amount of tension which this 
member carries as its share of the trussed stringer. 

In treating the web member of Fig. 18, it will be neces- 
sary to consider the form of the influence line for stress 
in MiLr,. Passing a section through U^Us,, M^L^, and 
L4L5, it is at once seen that for a load between the right 
abutment and L5, the stress in AI4L5 varies directly as the 
left-hand reaction. Similarly for a load between the left 
abutment and L4 the stress varies directly as the right- 
hand reaction. Consequently the criterion of Art. 17 is 
directly applicable to M4L5, the panel length in this case 
being the short panel L4L5. This reasoning applies in 
exactly the same way to the short panel length L3L4 in 
the case of UsM^ of Fig. 19. 

The maximum stress in the remaining portion of the 
member U3L5 in either of these two cases is then assumed 
to be that already determined for the other portion, in- 
creased or diminished, as the case may be, by the amount 
of stress which the part shown in dotted lines must carry 



lo6 INFLUENCE LINES. [Ch. II. 

in performing its duty as a part of the subordinate bracing. 
It is evident, however, that this result does not give the 
absolute maximum stress that may occur in UzM^ (Fig. i8)- 
or AI4L5 (Fig. 19). The stress in these members is com- 
posed of two parts — that part due to the true stress in the 
web member itself, and that part due simply to the weight 
carried by the tie-rod. A criterion indicating the position 
of the loading giving a simultaneous maximum condition 
of these two factors is not a simple one. As but a small 
error is involved in assuming that the position of the load 
causing maximum stress in one portion of the web member 
will also cause the maximum stress in another part, the 
same position of the loading is generally assumed for both 
portions of the web member. 

These investigations indicate that the maximum web 
stresses in trusses with subdivided panels may be obtained 
by means of methods previously deduced.* 



Art. 20. — Maximum Chord Stresses in Trusses with Subdivided 

Panels. 

Unloaded Chord. 

The stress in any unloaded chord member such as UsUi 
(Fig. 20) is found by passing a section through UsUi, U3M4, 
and L3L4, and with the centre of m.oments at L5, equating 
the moment of the external forces situated on one side 
of the section with the moment due to the chord stress. 
Although the centre of moments is at L5, the only external 
panel forces on one side of the section are Li to L3, the 
load at panel point L4 not being included. 

* The two cases of figs. i8 and 19 are recognized in order to complete the 
treatment, although it is doubtful whether an entire panel load can act as shown 
by the dotted lines of those figures. 



Art, 20. 



7RUSSES IVJTH SUBDIVIDED PANELS. 



107 



Tliis invalidates therefore the use of the simple criterion, 

IP' "/" 

deduced in Art. 6. By methods parallel to those of that article, 
however, the proper criterion may be quickly deduced. 

The influence line for the stress in U:iU4 is easily drawn, 
for as a load passes from the left end of the truss to panel 




Fig. 20. 

point L3, the stress varies directly as the right-hand truss 
reaction, i.e., as a straight line. A proper slope for this 
line maybe indicated (as in the case of moments. Art. 5) by 
laying off EL equal to I" on the left end of the reference line 
and connecting L with A , the right end of the reference line. 
In a similar manner, the line EF indicates that between 
Lo and L4 the stress also varies as a straight line, and its 
slope is found by laying off AK equal to /' at the left end 
of the reference line. 



io8 INFLUENCE LINES. [Ch. II. 

Since all influence lines for single loads are straight 
lines between panel points, the influence line may then 
be closed by the line JF. 

The notation which will be employed to develop the 
criterion is indicated in the figure. P, P", arid P' represent 
in general any load between Lo and L3, L3 and L4, and 
L4 and L12 respectively, while ni, n, and m' represent the 
corresponding influence ordinates. 

In general, then, the stress in'UsUi may be represented 
by the following expression: 

S=^IiP-m) + I{P"-n) + I{P'-m'). . . (i) 

The ordinate n may be divided by the line BF; then 

I(P"-n)^:iP"-n') + I{P"-n"). . . (2) 

By substituting ' this value of I{P" -n) in eq. (i) and 
by replacing the values of the influence ordinates by equiva- 
lent values found from similar triangles, the value of S 
becomes 

By moving the loads an infinitesimal amount to the 
left the change in the stress or AS becomes 






i\ P' 






CE 



Art. 20.1 TRUSSES WITH SUBDIVIDED PANELS. 109 

The condition for maximum stress requires that AS 
be placed equal to zero. By simplifying terms and by 
noting that 

there will be found that 

2'[(P + P' +P") -y] = S{P) + 2:[p" ■^. . (s) 

Eq. (5) is similar in form to eq. (5) of Art. 11, the latter 
being the criterion for finding the maximum bending 
moment at any point in any truss. It differs, however, 
in this essential, that in the present case the factor to be 
applied to the loads within the panel is p/q and not q/p. 

The position of the loading having been determined by 

means of this criterion, the stress in UsU^ is found by 

taking mom.ents about L5. The maximum stress in U^Us 

occurs with the same position of loading as UzU^, and 

♦ its value is also the same. 

Loaded Chord. 

The stress in any loaded chord member such as L3L4 
is found by passing a section through U3U4,, UsM^, and 
L3L4, and taking the centre of moments at U3. This 
permits the use of the simple moment criterion, 

IP I 
SP'~l' 



no INFLUENCE LINES. [Ch. IL 

of Art. 6, and no unusual conditions are encountered in 
finding the stress in L^L^. The maximum stress in L4L5 
is found with the same position of loading as L3L4, and 
the value of its stress is exactly the same. 



Art. 21. — Covinter-Stresses in a Vertical Post at an Angle 
in a Chord. 

Although the usual stress in the vertical post of a truss 
with inclined upper chords is compressive, it is possible 
for such a member to receive a counter tensile stress. 
This counter stress is not caused by the negative shear 
in a panel, but it occurs when the inclined web member, 
cutting the upper end of the vertical at the unloaded 
chord, sustains a small or even zero stress. In that case 
the equations of equilibrium, as applied to the unknown 
forces at the panel point in question, involve only the 
stresses in the two chord members and in the vertical post. 
Since the axes of the two chord members meeting at that 
point are not in a straight line, a component in the direc- 
tion of the post must result. In a through truss with 
horizontal lower chord this component furnishes tensile 
stress. It becomes necessary, therefore, in order to deter- 
mine this maximum tension in these posts to find that 
position of the loading which causes a zero stress in the 
inclined member; that is, the live load must cause a stress 
in this inclined member equal and opposite to that of the 
dead load. This position is easily determined by the use 
of the influence line. 

Assuming for the present that the position of the loading 
causing the maximum tension in the post is known, it 
becomes necessary to deduce an expression for the tensile 
stress in that member. Let Fig. 21 represent a truss in 
which it is desired to find the tension in the member U2L2 
when the stress in ^72-^3 is zero. It will be convenient 



Art. 21.1 COUNTER-STRESSES IN A VERTICAL POST. 



Ill 




112 INFLUENCE LINES. [Ch. IL 

first to determine the stress in U2L2 for a load P placed 
at panel point L2. The following notation will be used: 

p = panel length ; 

h=the height at the panel under consideration — in this 

case the length U2L2; 
di =the difference in the height of UiLi and U2L2; 
62 =the difference in the height of U2L2 and U3L3; 
d = d]_ — d2; 

Z=the length of truss; 
li =the length of the member U1U2; 
I2 =the length of the member U2U3; 
P =the load at the panel point L2 ; 
«=the number of panels (of equal length) between the 

left end of the truss and the member U2L2 ; 
Ci=the lever-arm of U1U2 about L2; 
C2=the lever-arm of [72^/3 about L2; 
a and a' =the angles of inclination of U1U2 and U2U3 with 

a horizontal respectively ; 
R =the reaction at left end of span. 
Then for the load P at L2, 

il-np)P 
^^~~l • 

The stress in U2L2 may be determined by finding, first, 
the stresses in U1U2 and t/2^3 and taking the difference 
of their vertical components as the stress in U2L2. By 
taking moments about L2 and omitting the stress in U2L3 
(which is assumed zero), the stresses in U1U2 and U2U3. 
will be 

C2 



Art. 21.] COUNTER-STRESSES IN A VERTICAL POST. HJ 

Their vertical components are then respectively 

R-np . ^ R-np . , 
■ sm a and sm a' . 

The tensile stress in ^72^2 is the difference of these 
vertical components: 

^^ -. „ /sin a sin a'\ 
V,U^R.np[^—-~^y . . . (I) 



But 



sm a: = -7— and sm a = 7- ; 



Also 



ci h , C2 h 

--=y- and —=7-. 
p h p h 



Therefore eq. (2) takes the form 

R-np-d R-n-d 



h-p h 



(3) 



Eq. (3) furnishes the expression by whose aid the in- 
fluence line for the tensile stress in the number U2L2 
may be drawn. As the stress in U2L2 varies directly as 
R the variation in stress for loads at other panel points 
may be represented by the ordinates between the base 
line AIN and the straight lines drawn from the end of the 
ordinate KL below L2 to M and A'', as in Fig. 22. It 
should be noted that this influence line, although for a 



114 INf-LUENCE LINES. [Ch. 11. 

web member, shows no reversal of stress throughout its 
length; it is to be used only when the stress in ^2-^3 is 
zero. The scale of Fig. 22 is found by the aid of eq. (3), 
which furnishes the value of the ordinate KL, the stress in 
U2L2, when the load P is placed at L2. 

In order to determine the position of the load which 
causes the maximum tension it is necessary to use the 
influence line for stress in the member f/2-^3. This line 
is represented by M'K'K"N' in Fig. 22, the method of 
deriving it being that of Art. 16. The loading must advance 
from the left toward the right until it is found by trial that 
the live-load stress in [/2-L3 is just equal to the dead-load 
stress found previously. The stress in U2L2 for this position 
of the loading may then be determined from the influence 
line drawn for that member. It is usual in the treatment 
of this counter-stress to substitute a uniform live load for 
the actual locomotive concentrations. The results thus 
obtained are not rigorously exact, but to use locomotive- 
wheel loads would involve needless refinement. 

One other point, however, remains to be considered, 
namely, that it may be possible to obtain a zero stress 
in the member U2L^ with a position of the loading that 
would cause greater tension in the member U2L2 than for 
the position just determined. This position of the loading 
is that caused by trains entering simultaneously upon the 
bridge from the two ends, but with a greater load on the 
left end than on the right end. In that case the stresses 
in U2LZ caused by the two different loadings tefid; to neu- 
tralize each other, but the left-hand loading must be so 
much greater as to balance the dead-load stress. Such a 
position causes greater tension in the member ^72^2, since 
according to the influence line for U2L2 every additional 
load on the truss increases its stress. 



Art. 21.] COUNTER-STRESSES IN A VERTICAL POST. 



IIS 



Example. 

As an example, the maximum tension in the member 
UsLs of the truss detailed in Chapter VI and shown in 
Fig. 23 will be determined, the locomotive concentrations 
being replaced by a uniform live load of 2500 pounds per 
linear foot of truss. 

For that case the dead-load stress in U^Mi has been 
found to be + 90,000 pounds ; therefore the live load must 




ElG. 24. 



12 Pane] lengths, all equaL29'2'=360V 

Loading taken uniform @ 2500 Lbs. per Linear Foott 



advance on the truss from the left end, until the stress 
caused by it in UzM^ is just equal to —90,000 pounds. 
The influence line for U3M4, is therefore drawn in the 
usual manner (Fig. 24), as follows: The chord member 
UzUi is continued until it intersects verticals through 
the abutments at a and b. Lines are then drawn from 
these points through the ends of the panel L3 and L5 to 
intersect at c. This point is projected to the base line AB 
to C, where it indicates the position of a load on the truss 
causing no stress in U3M4, (Art. 16). Placing a load of 



"6 INFLUENCE LINES. [Ch. II. 

unity at L3, it may then be found that the stress in U^M^ 
is —0.56; the influence ordinate FD is therefore drawn 
with that value. The completed line for UsM^ is then 
ADCEB. 

It will now be found by trial that the area ADC, 
which represents compressive stress, has the following 
value, if the length AC is entirely covered by the uniform 
load of 2500 pounds per linear foot: 

(AFxFD FCX FD\ /ACx FD\ 
25oo( + ^— j =25oo( ~ ) 



/ii6X.58\ 
-■ 2 500 ( I = — 8 1 , 200 pounds. 



It is seen that this stress does not quite equal the tension 
due to the dead load, but since its value is so nearly the 
same, and since uniform instead of concentrated loading 
has been used, it will be proper to assume that this position 
of the loading furnishes zero stress in U3M4. 

The influence line for U3L3 is then drawn. Placing a 
unit load at L3, eq. (3) will furnish the following value, 
remembering that 6^ = 4.5 feet, w = 3, /i = 54 feet, and i? =0.75: 

.75X3X4-5 , n^^ 
UzL3= — = +.1875. 

54 

The influence line for U^Lz is then AGB, the ordinate 
FG having a value of .1875. The tension in U3L3, for 
the loading just found, is therefore 

L''3L3 = 2 5oo (area ACF + area FGCH). 
Since 

CF = .i875 FC = 28.5feet 

AF = 87.5feet CH = . i6s 

.'. [/3L3 = +26,750 pounds. 



Art. 22.] CENTRE OF MOMENTS IVITHIN LIMITS OF TRUSS, u? 



Art. 22. — Influence Line for Stress in the Web Member of a 
Truss when the Centre of Moments Falls within the Limits 
of the Truss. 

The influence lines for stresses in web members deduced 
in the previous articles are applicable only to those trusses 
in which the centre of moments falls outside of the limits 
of the truss. In the case of the web member U1L2, shown 
in Fig. 25, the influence line may be determined in the 




Fig. 25. 

following manner. The stress in this member, U1L2, is 
found by passing a section through the members U1U2, 
U1L2, and L1L2 and dividing the sum of the moments of 
all the external forces (not shown) on one side of this 
section about c, the centre of moments by the lever-arm 
d of U1L2 about c. 



Ii8 INFLUENCE LINES. [Ch. II. 

: Assuming,, a load of .unity to pass between the left 
■end,,. of the truss, -/7o and the left end of the panel Ui cut 
by the section, the stress in the member may be ex- 
pressed as 

Td' 

in which- jc represents the distance from the left abut- 
meril;. Since the only variable in this expression is x 
the variation of the stress may be represented by a straight 
line, X being of the first degree. The stress is tensile. 
Similarly, for a load of unity between the right end abut- 
ment Uo and U2, the right end of the panel cut by the 
section, the variation in the stress may be represented 
by the straight line 

x-h 

Td' 

assuming x now .to be the distance from the right abut- 
ment. The stress is again tensile. 

These two expressions are proportional to I2 and h 
respectively. Erecting MB at the left end of the reference 
line MN equal to h and NA at the right end equal to I2 
and connecting B and A to N and M respectively, the 
lines ME and DN will form part of the desired influence 
line. Since it has been shown (Art. 10) that influence 
lines between panel points are straight lines, it then only 
becomes necessary to connect the points E and D hj a, 
.straight line to complete the diagram. 

It is therefore seen that in the case of a web member 
of the kind shown in the figure, counter-stress never occurs. 
The member sustains its maximum stress with the entire 
truss covered with load. A similar construction might 
be developed for the vertical members. It is possible to 
deduce a criterion for maximum stress in the member 



A.RT. 23.J INFLUENCE LINES FOR SKEIV BRIDGES. 119 

U1L2 by means of the influence line shown in the diagram, 
but since this form of truss is rarely built such a criterioh 
need not be developed. The treatment of an equivalent 
xiniform load is, moreover, so simple that it should be used 
to replace the locomotive concentrations. For that char- 
acter of loading it is only necessary to multiply the area 
of the figure MEDCN by the proper intensity of the 
uniform load to obtain the maximum stress. 



Art. 23. — Influence Lines for Skew Bridges. 

Skew bridges are trussed structures in which the ends 
of the pair of trusses forming the bridge do not lie in a 
line perpendicular to its axis. The skew of a bridge is 
the distance between the projections of the ends of the 
trusses measured on the axis of the bridge. This distance 
is not necessarily the same at the two ends. In Fig. 28, 
which represents the plan of the lower chord system of a 
skew bridge, the skew at the left end of the bridge is a 
full panel length, or LqLi, while at the right end of the 
bridge it is eL^ == cd. This example illustrates a skew bridge 
in the most general form. 

It is customary to place the floor -beams, supporting the 
floor system, at right angles to the trusses, as shown in the 
figure, and for convenience in the manufacture of the 
details of the portal bracing it is preferable to design the 
end posts of opposite trusses with the same inclination. 
This end is attained by moving the upper end of the end 
post of the further truss, that is, U &' in Fig. 27, half the 
skew distance back of the vertical projection of Le', while 
Uq, Fig. 29, is moved half the skew distance in advance of 
the vertical projection of Lq. The members U^'L^, 
Fig. 27, and U&Lt, Fig. 29, will then be parallel. The 



I20 



INFLUENCE LINES. 



[Ch. II. 



upper chord members TJ'^JJ^ and V^V^ will not be equal 
in length, nor will Uq'Lq and U^L^ be either parallel or 




L'3 L'4 L'i L'ol L'; 

L'l Fig. 27. ELEVATION OF FAR TRUSS. l'o L'- 



FlG. 28. PLAN OF LOWER CHORDS. L,. 

U5 U3 U4 Ur, U,. 




FlG.29. lELEVATION OF NEAR TRUSS. I 
I I 




INFLUENCE UNE 
FOR U3 L4 ■ 



vertical ; but this is preferable to having the end posts non- 
parallel. 



Art. 23.] INFLUENCE LINES FOR SKEIV BRIDGES. 121 

At the left end of the bridge, such construction in the 
example considered is not necessary, for the skew distance 
is a full panel length. 

Should the skew distances at the two ends of the bridge 
be equal, the two trusses would be exactly alike except 
that they would be turned end for end. 

The moving load is taken to pass along the centre line 
ad of the bridge, and this necessitates changes in the in- 
fluence lines for stresses in the various members. Treating 
first the member U2UZ of the truss of Fig. 29, it is seen 
that its stress will vary as the moment at L3. If the 
tmss Li . . . L7 carried loads on the span- Li . . . L7, the 
moment influence line at L3 would be MQN, Fig. 30. 
By referring to Fig. 28, however, it will be seen that loads 
between Li and Le only are carried on that span length. 
Loads are carried on the line cd only in the panel LqLt, 
and a load at d causes no stress in the truss L1L7. Since 
the influence line is straight between all panel points, the 
line SK, Fig. 30, will exhibit the variation of the moment 
at L3 for loads between c and d. The final influence line 
is therefore MQSKN. For uniform loading it is only nec- 
essary to measure the area shown shaded, and to multiply 
by the proper intensity of the loading to obtain the value 
of the moment. For concentrated loads the criterion 
previously deduced would require modification for the 
portion SK of the line, although ordinarily no serious 
inaccuracy results from the use of the general formula. 
The values of the moments for different positions of such 
concentrations may also be found by trial and comparison. 

The treatment of chord member LqLy or UsUq is shown 
in Fig. 31. If the truss had the span length L1L7, the 
influence line would be MSN, the centre of moments being 
at Ue ; but owing to the skew, the final line becomes MQK. 

It should be noted that in this truss the stress in L^Lq 
is not equal to the stress in L^Lj if there be a load at Lq, for 



122 INFLUENCE LINES. [Ch. II. 

the member UeLe adds a horizontal component at panel 
point Lq. 

The web members are also treated in precisely the same 
way. Fig. 32 illustrates the treatment of U3L4, for which 
the final influence area is shown shaded. 

The treatment of the further truss, Fig. 27, does not 
differ from the preceding. The influence line for the 
member Lo'L/ is shown in Fig. 26. If the truss had the 
span length LqLt , the influence line would be MQX, 
M and X being vertically above the ends Lq and L^ of 
the truss, and Q being above the centre of moments. The 
loads advance along the centre line ad, however, and those 
at a and d cause zero stress in the truss. The true in- 
fluence line for Lq'Li is therefore TQSN, Fig. 26, and 
the influence area for uniform loading is shown shaded. 

The preceding treatment is entirely general, and may 
be applied to any skew bridge. 

Art. 24. — Influence Lines for Double-intersection Trusses. 

The influence line possesses distinct advantage in the 
treatment of double-intersection trusses whether with parallel 
or broken chords, although in the example treated, Fig. 33, 
a truss with parallel chords only will be considered. The 
usual analytical method of treatment is to assume that 
the structure is composed , of two systems of trussing, 
Figs. 34 and 35, acting independently. The maximum 
stresses that can occur in each system are found by trial, 
and if the same member acts in both systems, the sum of 
the stresses found is taken, provided both have been found 
for the same position of loading. Conditions of loading 
which fail to cause simultaneous stresses in the two sys- 
tems must not be considered in such summations of 
stresses. 

In the treatment of chord members the entire structure 



Art. 24-1 TREATMENT OF DOUBLE-INTERSECTION TRUSSES. 123 

is covered by load, and its position must not be changed 
when considering the same member for the two systems. 

In the graphical treatment it is only necessary to con- 
sider that a load at a panel point is carried entirely by 
the system of which that panel point is a part. If the 
load is within a panel, it is distributed in some manner 
between the two systems. 

The truss shown in Fig. t,t, contains an even number 
of panels and avoids a possible ambiguity which might 
be caused in two systems of trussing not symmetrical about 
a vertical centre line, i.e., equal loads placed at equal dis- 
tances from the centre of span would rest on two different 
systems. The treatment by influence lines avoids this 
ambiguity. 

Chord Members. 

Let it be required to find the influence line for stress in 
the member U4U5, Fig. 33 . This member forms part of U4Ue 
of Fig. 34 and part of UzU^ of Fig. 35. The influence 
line for U4UQ is MRN of Fig. 36, the point R being the 
projection of the centre of moments Le- Similarly, MSN is 
the influence line for U^U b, the centre of moments being 
at L5. It has been assumed that a load at a panel point 
acts only in the system to which that panel point belongs. 
A load at L4, then, causes a stress which may be repre- 
sented by the ordinate below it enclosed by the line MRN ; 
and this applies to all the lower panel points of Fig. 34. 
A load at Li is, however, taken as divided equally between 
the two systems of trussing. Similarly, loads at the lower 
panel points of Fig. 35 causes tresses represented by ordi- 
nates enclosed within the line MSN. But all influence 
lines between panel points are straight lines. Hence the final 
influence line for U ^U ^ is represented by MabcdSRghijkN , 
and for uniform loading the stress would be obtained by 
multiplying the area shown shaded by the proper intensity 
of loading. For reasons already explained, the points a 



124 



INFLUENCE LINES. 



[Ch. II. 



and k lie midway between the two original influence lines 
drawn. It is evident that for chord stresses no serious 



U, U» Us Ui Us Uo 




error is involved, if the area of either of the first two in- 
fluence lines be measured. The choice must, however, 
be left to the designer's judgment. 



Art. 24.] TREATMENT OF DOUBLE-INTERSECTION TRUSSES. 125 

Web Members. 

The treatment of web members follows precisely that 
of chord members. In Fig. 37 the line M'QR'N' represents 
the influence line for the member U^Lq, treating it only 
as a member of the system of trussing shown in Fig. 34. 
The points Q and R' are projections of the panel points 
L4 and Le. Since a load at a panel point of the other 
system of trussing causes no stress in U ^Lq, the projections 
of those panel points on the base line M'N' all represent 
zero stresses; that is, the base line M'N' is the influence 
line for stress in U4LQ, when the loads are placed at the 
panel points of Fig. 35. The final influence line is therefore 
M'a'b'c'Q'e'R'g'h'i'j'k'N'. The points k' and a' lie, as 
before, midway between the two influence lines. 



CHAPTER III. 

THE THREE-HINGED ARCH. 

Art. I. — To Pass a Funicular Polygon Through Three Points. 

Let it be required to pass a funicular polygon through 
the three points A, B, and C, Fig. i, for the loads Pi, 
P 2 ■ ■ ■ Pi shown. The loads Pi . . . P3, situated between 
the points A and B, will be considered separately from 
the loads P^ . . ■ P7 situated betw^een B and C. Any other 
division of the loads might be made, but it will usually 
be most convenient to have the centre point B divide 
the loads. 

In Fig. 2 the force polygon P1P2P3 is drawn, and the 
closing line furnishes the value of a resultant which will 
balance the loads. Choosing any point, such as 0', as a 
pole, the rays i, 2, 3, and 4 are drawn and transferred 
to Fig. I to form the funicular polygon i, 2, 3, and 4. 
Through the points A and B of that figure lines R and Ri 
parallel to the resultant of Pi, P2, P3 are first drawn. The 
intersections of the bars i and 4 with those lines then 
determine the bar 5. The direction of this bar is then 
transferred to Fig. 2 and drawn through the pole 0', thus 
determining the values of the reactions R and Ri acting 
at the points A and B, respectively, for the loads between 
A and B. 

Similarly, choosing as a pole 0", the reactions R2 and 

R3 for the loads between B and C are found for the points 

B and C. The conditions of the problem require the 

126 



Art. 1. 1 FUNICULAR POLYGON THROUGH THREE POINTS. 



127 



funicular polygon i, 2 ... 5 to pass through the points 
A and B ; that is, the bar 5 should be the line 5' connecting 
A and B. Transferring therefore the direction of 5' to 
Fig. 2 and drawing it through the point m, it will be evident 



Fig. I. 




\-^r-]f}^¥f/r^X 





Fig. 2. 



that any point on this line will furnish a funicular polygon 
for the loads P\ . . . Pz which will pass through A and B. 
In precisely the same way, then, the line 11' may be drawn 
in Fig. 2 for the forces P4 . . . P7, 11' being parallel to the 
line connecting the points B and C. Any point on this line 
will furnish a pole for a funicular polygon passing through 
the points B and C and constructed for the forces situated 
between them. Consequently, the intersection of the 
lines 5' and 11' will furnish the only pole which will pro- 
vide a funicular polygon passing through the three points 



128 THE THREE-HINGED ARCH. [Ch. III. 

A, B, and C for all the forces shown. This polygon is 
not shown in the figure, but after the pole O is foiuid its 
construction is obvious. 

The rays 12 and 13, being the extreme rays for the true 
polygon, represent bars in a framework holding in equilib- 
rium the given forces Pi . . . P7. They pass through the 
points A and C respectively, and, as will presently be seen, 
are the reactions for a three-hinged arch having hinges 
at A, B, and C and holding in equilibrium the forces given. 

Art. 2. — Determination of the Reactions of a Three-hinged 

Arch. 

An arch provided with three hinges is a structure 
which is statically determinate and in the treatment of 
the stresses in its members involves only principles already 
established. 

Fig. 3 illustrates a three-hinged spandrel-braced arch 
provided with hinges at each abutment and at the crown, 
panel point 13. It is desired to find the stresses in the 
structure for a fixed position of the loading. 

The dead loading, carried at the upper chord panel 
points, and assumed to have a value of 28,300 pounds 
per panel per truss, will be treated. The end-panel loads 
are each taken at 14,150 pounds only. The supporting 
reactions at the abutments for this type of truss are no 
longer vertical. Their points of application are given, 
but their directions are imknown. Since they hold in 
equilibrium the loading on the truss, the problem becomes 
that of determining the amounts and directions of two 
forces whose points of application, i and 14, are given, 
and which hold in equilibrium a known set of forces. The 
problem would evidently be statically indeterminate ex- 
cept for the condition imposed by the centre hinge, which 
is that the sum of the moments of the external forces on 



Art. 2.] 



REACTION OF A THREE-HINGED ARCH. 



129 



either side of the centre hinge must equal zero about that 
point. The solution of the problem, then, consists in 
passing a funicular polygon through the three hinges 
(see Art. i), the pole distance of that polygon representing 
the horizontal component of each reaction. This follows, 




Three Hinged Spandrel Braced.Arch 
12 Panels at 16'8"=200'.0' 

Rise=10'0' 
Depth at Ciown.= sV 




Fig. 3. 



Scale In Pounds 



Fig. 4. 



since a funicular polygon represents bars in a structure 
which carry direct stress only for the given system of loads, 
and if a bar of this fvinicular polygon passes through any 



13° THE THREE-HINGED ARCH. [Ch. Ill- 

given point, such as a hinge, there will be no bending mo- 
ment at that point. Therefore a funicular polygon through 
the three hinges will furnish the values of the stresses in 
the supporting or end. bars, which are the reactions. 

In Fig. 4, therefore, 0' is chosen arbitrarily as any pole 
for the loads on the left half of the structure. The polygon 
2, 3, 4 ... 9 in Fig. 3 is then drawn, and the last ray, viz., 
9, transferred to the force diagram, Fig. 4. This deter- 
mines the point x in the line of the loads, from which point 
the line xO is drawn parallel to the line connecting the 
two hinges 13 and 14. Any point on this line xO will 
furnish a pole from which a funicular polygon passing 
through the points 13 and 14 may be obtained. 

Since the loading is symmetrical about the centre it 
will not be necessary to complete the graphical operations 
for the right half of the structure, but the line x'O may 
at once be drawn parallel to the line connecting the hinges 
I and 13, w being situated midway between x and x' . 
The intersection of xO and x'O determines at once the true 
pole of a frmicular polygon which will pass through the 
three points i, 13, and 14 of the arch. The pole distance 
is, then, Ow, -and it represents a horizontal component in 
the reactions of 312,000 pounds. The reactions themselves 
are then compounded from this horizontal force and the 
vertical reactions aw and a'w; they are graphically repre- 
sented by Oa and Oa' , and are each equal to 398,000 pounds. 

The horizontal component H may perhaps be more 
quickly determined by analytical methods, for by taking 
moments about the centre hinge 13 the following equation 
will result : 

HX40 = 242,950 X 100— 14,150 X 100 

-4i,6oo[i6f(5-l-4 + 3-f-2-|-i)]; 

.•. if = 312,000 pounds. 
This value checks that found graphically. 



ART. 3-] MOMENTS IN THREE-HINGED ARCHES. 131 

After the reactions have been determined, the problem 
presents no further difficulties so far as the determination 
of the stresses in the members is concerned. The stresses 
in the members m.eeting at panel point i are first to be 
found, then those meeting at panel point 2 are next to 
be treated, and the. other panel points in the numerical 
order shown in the diagram. 

The treatment of the stresses has thus far involved a 
fixed position of the loading, but the positions of the load- 
ing causing the greatest stresses in the various mem.bers 
may easily be determined by the aid of the influence lines, 
in the case of three-hinged roof-trusses the graphical 
methods thus far outlined are sufficient to find all the 
desired stresses, for in such cases the loading, in whatever 
manner it may be applied, is always fixed in position. 



Art. 3. — Moments in Three-hinged Arches. 

The graphic determination of the reactions of a three- 
hinged arch, as found in the previous article, is not as 
simple as the anal5rtic determination, and hence the treat- 
ment of a three-hinged arch by the combination of algebraic 
and graphic methods is more convenient than by the 
graphic process alone. 

The application of influence lines to these structures 
is of distinct value, since by their aid the positions of 
loading causing maximum stresses in the various members 
may be at once determined. For this class of structure, 
the use of wheel-load concentrations constitutes, however, 
unnecessary refinement and such loading will not be con- 
sidered in detail. It is sufficient to state again that if the 
influence lines for stresses have been drawn for a unit load 
P, the maximum stress caused by any series of concentra- 
tions may be easily determined by trial. The general 



132 



THE THREE-HINGED ARCH. 



[Ch. III. 



value of the stress for a series of loads would then be 
represented by IP ■ n, where P represents the value of the 
load at any poitit and n the corresponding ordinate. 

Before considering influence lines for three-hinged arches 
it will be necessary to obtain a general expression for the 




Fig. 6. 



bending moment at any section of a three-hinged arch 
for any system of loading. Let Fig. 5 represent an arch 
hinged at the points A, B, and C, and let the bending 
moment at the point X be required. 

The reactions R and Ri may be found by means of the 
funicular polygon, their values corresponding respectively 
to the rays i and 7 in the force diagram. The horizontal 
and vertical components of these reactions are represented 
by H, V, and Fi respectively. Taking moments of all 
the left-hand external forces about X, the general expression 
for M becomes 

M = IV-x + H-y (i) 

where V represents the vertical components both of the 
reaction and of the given loads, x their respective lever- 



Art. 4.] INFLUENCE LINES FOR REACTIONS. ^ 33 

arms, H the horizontal component of the reaction, and y 
its lever-arm. But it is evident that the first term in the 
right-hand member represents the bending moment due 
to vertical loading at the section X for a simple non-con- 
tinuous beam of span AC, and its graphical representation 
is the product of the intercept YZ by the pole distance 
H or y'-H. The second term in the left-hand member 
of eq. (i) is the product of the same pole distance by a 
different intercept, viz., XZ=y. The bending moment at 
any section of the arch is, therefore, graphically repre- 
sented by H{y — y'), or the product of the pole distance, 
H, by the vertical intercept between the point taken 
on the arch and the true funicular polygon. It is then 
seen that the line of the arch for the figure shown is sub- 
jected at its various points to positive moments for the 
right-hand section, BC, and negative moments for the 
left section, AB. At the centre hinge B, the moment 
reduces to zero, as it should. 

It will be found that this graphical representation of 
the moments to which an arch may be subjected is of 
value in determining the stresses in arches differently 
conditioned, such as arches with fixed ends or arches 
with two hinges. 



Art. 4. — Influence Lines for Reactions in Three-hinged Arches. 

Let it be required to determine the variation of the 
horizontal and vertical components of the reactions at A 
and C as any load P passes over the three-hinged arch 
shown in Fig. 7. 

The horizontal projection of the two parts of the arch 
are represented by /i and 1 2, while the total span is /. The 
rise of the centre hinge above the horizontal line A and C 
is h. 



134 



THE THREE-HINGED ARCH. 



[Ch. III. 



Taking B as the origin of coordinates, and measuring x, 
which is the distance in either direction of any load P, 




R, Fig. 7. 



Fig. 8. 



Fig. 9 



from B, the horizontal component H of the reaction at 
A, for the load between A and B, is 



H = 



h 



(i) 



V ^ represents the vertical component of the reaction at C. 
But 

P{U-x) 



V = 



I 



and therefore 



H = 



P{k-X)l2 
I 



(2) 



This expression may be represented graphically by a 
straight line, for the variable x appears in the first power 
only. 

Erecting, then, in Fig. 8, on the base line ST a vertical 
TX below the hinge C equal to I2 and connecting X with 



Art. 4.] INFLUENCE LINES FOR REACTIONS. I35 

5, it will be found that the portion of A'5, viz., WS, 
included between vertical lines through A and B, will 
represent the- influence line for H when the load lies be- 
tween A and B. For by similar triangles the ordinate 

{I I — x)lo 
directly below the load is equal to -, "-. Hence the 

line WS is the influence line for a unit load. 

In the same manner the line WT may be drawn to 
complete the diagram, the ordinate SY being erected 
below A eqvial to l\ and the line YT then being drawn. 

The final influence line for H is therefore SWT, and' 
it obtains its maximum value for a load at the centre 
hinge, i.e., 

_ P(/l/2) 
■'^ max 7 

The influence lines for the vertical components Va and 
V ^ dift'er in nojway from those found for simple beams, 
in consequence of the statical conditions of the structure. 

In Fig. g, therefore, the line T'U' represents the in- 
fluence line for Va, and a similar line sloping in the opposite 
direction would be used for V ^. 

It has been shown in Figs. 8 and 9 that each of the 
rectangular components of each reaction varies as the ordi- 
nates of a straight line ; consequently their resultant, or the 
reaction itself, must vary in precisely the same manner. 
Hence, as the load P passes over the bridge, the variation 
in the value of the reaction R may also be represented 
by a straight line. The slope of such a line is a question 
of no importance; it suffices to know the law of vari- 
ation. 



136 



THE THREE-HINGED ARCH. 



[Ch. IIL 



Art. 5, — Influence Line for Stress in any Chord Member of 
a Three-hinged Arch. 

Let Fig. 10 represent a three-hinged spandrel-braced 
arch with a horizontal upper chord and with vertical and 
inclined web members. The treatment of the problem, 
although applied to this particular truss, will be given 
in the most general manner so as to apply to any other 
form of three-hinged arch. 



Center of Moments 




Let it be desired to find the influence line for stress in 
the chord member UiUz- Passing a section through the 
members t/2^3, U2L3, and L2L3, and taking the centre of 
moments at L3, the stress in U2U3 will evidently vary at 
precisely the same rate as the moment of the external 
forces taken about the point L3. The influence line for 
stress in U2U3 may, therefore, be represented by the moment 
influence line for the point L3. It has already been shown 
(Art. 3) that the moment at any point in a three-hinged 
arch is represented by the product of the pole distance 
by the vertical intercept between the centre of moments 



Art. 5-] INFi.UENCE LINES FOR CHORD MEMBERS. 137 

on the arch and the true funicular polygon. For the truss 
shown, such a variable moment I\I for a single, unit load 
passing along the structure is represented by 

M=M,~H-y, (i) 

where M^ is the variable moment for the load passing over 
a simple' beam with a span length LqLo, and where the 
product H ■}' is the moment due to the horizontal compo- 
nent of the reaction, y evidently being constant. On the 
base line ST (Fig. ii) the influence line for M^ may then 
be drawn in the usual manner, by erecting at 5 an ordinate 
SS', equal to the distance /' between the centre of moments 
Ls and the left-hand abutment Lq. and connecting S' with 
T. Performing a similar operation for the right point of 
support, the influence line M^ is represented by SGT. 
From this line there is to be subtracted at every point 
the moment H-y, which may be found in the following 
manner: 

If the line of action of the reaction at Lq passes through 
Ls, the centre of moments, and if the load causing this 
reaction should be placed to the right of the section passing 
through the members cut, the resultant moment about L3 
will be equal to zero ; for the only force to the left of the 
section will be the reaction and its lever -arm about the 
centre of moments is zero. Such a position of the load 
causing no stress is found at the point Y, which is the 
intersection of lines drawn through Lq and L3, and Lq 
and L5; for the polygon LqYLo is the funicular polygon 
for this one position of the load, since it passes through the 
three hinges. This position of the load, causing no stress 
in some particular member, will in future always be repre- 
sented by Y. Projecting this point vertically downward 
until it intersects GT at / will fix that point on the final 
influence line, at which the stress is zero. 



138 THE THREE-HINGED ARCH. [Ch. III. 

It has been shown in the previous article that the hori- 
zontal component of the reaction H varies as the ordinates 
of a triangle, the apex of the triangle being directly below 
the centre hinge. In eq. (i), under consideration, the prod- 
uct of H and y is required, but since y is constant the 
product will vary in the same manner as H\ that is, as 
straight lines between the end and centre hinges. Three 
points of the variable H-y are known, viz., 5, T, and /. 
The lines SJK and KT may then at once be drawn to 
represent the variation of H-y, and any ordinate in the 
areas SGJ and JKT will represent the moment for the 
member U2U3 for the load placed over such ordinate; 
for, according to eq. (i), that ordinate expresses the re- 
lation M^ — H-y. The shaded portion to the right of the 
point Y will represent negative moments, and the other 
shaded portion positive moments. For maximum stress of 
one kind the structure must then be covered with load 
from the point Lq to Y, and for maximum stress of the 
opposite kind from Lq to Y. The magnitude of the stress 
may at once be determined if the scale of the influence line 
is known; and this scale may be found if the stress in 
the member U2U3 be known for the load at any position, 
such as at U3. 

If, therefore, an ordinary force diagram be drawn for 
the entire structure for a single load P at any point, such 
as Us, the stress in every member of the structure will 
be determined for this position of the load, and consequently 
also the scale of every influence line. In the present 
instance, for the member U2U3, the force diagram would 
furnish the true value of GG', and consequently the scale 
for any other ordinate in the area SGJ or JKT. It is 
then only necessary to multiply the area of the triangles 
SGJ and JKT by the proper intensity of loading to obtain 
the final maximum stresses for U2U3. This method of 
determining the scale of any influence line by means of a 



Art, 5.] 



INFLUENCE LINES FOR CHORD MEMBERS. 



139 



force diagram drawn for a single load placed at any one 
fixed position will be applied to the case 'Of web members, 
and it will not be necessary to repeat this explanation for 
those members. 

The treatment of the stress in a chord member, such as 
U5 Ue, Fig. 12, for which the centre of moments, L5, is not 
vertically below a panel point, follows precisely the same 
methods; but since the influence line between the panel 




Fig. 12 



points U5 and Ue is a straight line, it becomes necessary 
to remove the angle BCD of the influence line ACE, pre- 
cisely as in the case of the simple truss of Art. 11, Chap. 11. 
The influence line ACE is, therefore, drawn considering a 
point vertically above L5 as the centre of moments; the 
intersections of verticals drawn from the ends U5 and Uq 
of the panel to AC and CE determine the points B and D. 
Between B and D the influence line must be a straight 
line, and the final line is therefore ABDE. 



140 



IHE IHKtt-HlNGED ARCH. 



[Ch. UL 



Art. 6. — Influence Line for Stress in any Web Member of a 
Three-hinged Arch, 

Let it be desired to find the influence line for stress in 
the member U1L2, Fig. 13, as a single load P passes along 
the structure. It may easily be shown that as long as 
the load P is to the left of the panel cut by the section 
passed through the members U1U2, U1L2, and L1L2, the 



Fig. 13. 



Fig. 14. 



I Center ot lloments , I 

Up U, U,^ U3 ^■'jXU.'. U4 U3 U'a U'l 




variation in the stress may be represented by a straight line. 
For the only external force to the right of the section is 
the right-hand reaction, which varies as a straight line for 
a single load, and as the moment of the reaction is taken 
about X, the intersection of U1U2 and L1L2, the variation 
in the stress of U1L2 for a load to the left of the section 
may also be represented by a straight line. Similarly, 
for the load on the right half of the arch between U2 and 
Ud, the stress varies directly as the left-hand reaction, and 
may be represented by straight lines intersecting, in the 
vertical dropped from the hinge. This variation is similar 
to that of the left-hand reaction. 

As the load passes over the panel U1U2, through which 
the section has been taken, the variation of the stress may 
be represented by a straight line, according to the treat- 



Art. 0.] INFLUENCE LINES FOR IVEB MEMBERS. 141 

merit of Art. 10, Chap. II. Four straight lines arc, there- 
fore, required in determining the influence line for stress 
in web members. 

The stress in U\L2\s zero for a load placed at the point 
Y, this point being determined in precisely the same 
manner as in the case for chord members, i.e., it is the 
intersection of a line connecting the points Lq and L5 
with the line connecting Lq with X, the centre of moments. 
This point Y is projected vertically downward on the base 
line ST to G, and furnishes one point of the influence line. 

It has been shown (Art. 16, Chap. II), in the case 
of the influence line for web members for simple non- 
continuous trusses, that the variation of the slopes of the 
positive and negative portions of those lines may be found 
by drawing lines from any point projected vertically below 
the centre of moments X to the ends of the span. This 
rule applies also to that half, U0U5, of the structure 
shown, of which the web member in question is a part, as 
the following analysis will show. 

The origin of coordinates from which x, the position of 
the load P at any instant, is measured is at X, the centre 
of moments, and this point divides the span / into the 
two portions ti = UoX and 12=XUq. The vertical dis- 
tance of X, the centre of moments, above the line connecting 
Lo with Lq is y and the corresponding distance of the centre 
hinge above this line is h. The vertical components of 
the left- and right-hand reactions at Lo and Lq are Vi and 
V2, and their horizontal component is H. 

Then for a load of iinity between Uo and Ui, the 
moment of the stress Si for U1L2 becomes, by taking 
moments about X of the external forces to the right of 
the section, 

Si-g = V2l2-Hy, 

g being the lever-arm of U1L2 about X. 



142 THE THREE-HINGED ARCH. [Ch. III. 

l\—X V9I9 

But "^2=^^ and H=^. 

Iherefore Sx-g = ^ -^ (i) 

Similarly, for a load unity between U 2 and t/s, the 
moment of the stress 52 becomes 

S2-g = y\l\-l-iy\ 

, , .. {l2±X)_ . „ Fl/l 

out K 1 = ; — and H = , . 

rr.^ f ^ (l2Xx)h (l2 + x)h-y . , 

Iheretore 52-g = 1 — v, (2) 

Both equations (i) and (2) represent straight lines, since 
the variable x is of the first degree only. When x = o, 

c cr ^1^2/ y 

g-l\ h 

that is, these two lines intersect in a point vertically below 
the origin of coordinates X. 

In Fig. 14, therefore, A is any point below X, and the 
two lines ^45 and AB are drawn to represent portions of 
the influence line, for a load at S causes no stress in U1L2, 
nor, as just explained, does a load at G. Again, the lines 
passing through S and G must intersect on the vertical 
below X. 

The remainder of the influence line may then be drawn 
from the conclusions reached in the beginning of this 
article. The line BA must be continued as a straight line 
to C vertically under the centre hinge, for the variation of 
stress in U1L2 is a straight line for the portion of the span 
U2U5; SD is the influence line for a load between Uq 
and Ui ; DB that for a load between Ui and U2, and CT 
that for the portion of the span U^Uq. The shaded areas 



Art. 6.] 



INFLUENCE LINES FOR IVEB MEMBERS. 



143 



above the base line ST indicate stress of one kind and 
show not only what portion of the span must be covered 
with uniform load, but also the magnitude of the stress 
caused by that loading, for its value may be found by 
multiplying the positive shaded area by the intensity of 
the loading. As before the scale of the influence line may 
be foUnd by drawing a stress diagram for the entire arch 
for one load at any point and cornparing the value of the 
stress for this load with the ordinate below it on the in- 
fluence line. 

The shaded area below the base line indicates the 
portion of the structure to be covered by the load to cause 
negative stress, and its value may be calculated as just 
indicated. 

Figs. 15 and 16 illustrate the influence line for stress 



X-Ceuter of Moments 

, 1 / , , , 

Up Ui Uo U:)j U4 Us Ui Ua j/U- ^^ '^n 



Fig. 15. 



Fig. 16 




in the web member U3L4 of the same three-hinged arch, but 
for this member the centre of moments X falls on the 
opposite side of the centre hinge, as does the point Y, the 
position of the load causing no stress. This influence line 
requires a slight additional explanation. As before, the 
influence lines for the portions U0U3 and U^Us of the 
truss must intersect at A, and they must pass through 
5 and G respectively, G being the projection of Y on ST. 
They are shown in Fig. 15 as SD and BG, but since the 



144 THE THREE-HINGED ARCH. [Ch. III. 

influence line for the portion U5U0 is a continuous str'aight 
line, the point C where BG cuts the vertical through the 
centre hinge must be connected to T. The final influence 
line is, therefore, SDBCT, and it is at once evident what 
portions of the structure must be covered by the load to 
cause the maximum stresses of opposite kinds. 

If the web members are alternately inclined, as in the 
form of webbing adopted in the Warren truss, the influence 
line suffers for that panel length of which the web member 
is a part, the modification explained in Art. 11, Chap. II; 
this change is also precisely the same applied to the chord 
members in Art. 5 of this chapter. 



CHAPTER IV. 
CANTILEVERS. 

Art. I. — Definitions. 

A BEAM continuous over n points of support (Fig. i) 
of which n—i points are rolHng supports, permitting 
longitudinal motion, requires the determination of n reac- 
tions. For the case of vertical loads only and iinder the 
condition that the rolling points of support move without 
friction on horizontal surfaces, two equations of condition 
only can be applied, namely, HV =o and I^M = o, since the 
third equation of condition, HH^o, disappears. There 
are required, therefore, for finding all the reactions* w — 2 
other equations of condition, if the problem is to be stat- 
ically determinate. Such equations are afforded by the 
insertion of n — 2 hinges, in the structure between the 
extreme points of support. Every hinge corresponds to 
an equation expressing the condition that the sum of the 
moments of the external forces on one side of it and about 
that point must be equal to zero. It should be observed, 
however, that without disturbing equilibrium two hinges 
only may be inserted between two adjacent points of 
support. A continuous beam may, therefore, be made 
statically determinate by the insertion of n—2 hinges. 
Such structures are termed cantilevers. 

Fig. I illustrates a cantilever having four points of 
support and two hinges, at A and B. The span AB repre- 
sents then a beam simply supported at its ends. The 

145 



146 CANTILEyERS. [Ch, IV. 

Spans CD and EF are termed anchor-arms and the por- 
tions DA and BE cantilever-arms. Other types of canti- 
levers may be constructed in which the hinges A and B 
occur in the spans CD and EF respectively. Such forms 
will not be analyzed separately, but the analysis deduced 
from Fig. i will be of such general character that it may 
be easily applied to any other form. 



Art. 2. — Bending Moments in Cantilevers. 

The funicular polygon is applied to a cantilever for 
the purpose of determining the moments at the various 
sections. of the structure. Let the loads Pi, P2, P3 . . . Pn 
(Fig. i) act on the various spans, as shown. Fig. 3 shows 
these loads laid down in regular order, with the poles 
0', 0", 0'", chosen arbitrarily, but all with the same 
pole distance. One pole might have been chosen, but it 
will prove more convenient to choose separate poles for 
the loads on the different span lengths CD, DE, and EF. 
The funicular polygon (Fig. 2) is then constructed in the 
usual manner with the rays i, 2 ... 14, but no closing 
lines can as yet be determined. The positions of the closing 
lines may be found immediately, however, since directly 
below the points A and B there are points of zero bending. 
The bar 16 is, therefore, drawn in Fig. 2 so as to show 
zero moments below A and B. At the points where bar 16 
intersects the lines of the reactions R2 and P3, bars 15 
and 17 may then be drawn to the reactions Pi and P4 
respectively, showing zero moments below points C and F. 
The shaded portions of the funicular polygon (Fig. 2) 
represent, then, in the usual manner the bending moments 
at the various sections of the structure for the fixed position 
of the loading shown. 

The magnitudes of the reactions Pi, P2, P3, and P4 



Art. 



REACTION INFLUENCE LINES FOR CANTILEVERS. 



147 



may also be found from Fig. 3, it being only necessary 
to insert the three closing rays 15, 16, and 17 through 




|A '17/14 

>l^^io. 3. 

the poles 0', 0" , and 0'" respectively. The brackets 
enclosing portions of the load line furnish the desired 
quantities. 



Art. 3. — Reaction Influence Lines for Cantilevers. 

Fig. 4 illustrates a general type of cantilever structure 
in which the hinges are at A and B, and for which it is 
desired to represent the variations of the reactions Rx, R2, 
R3, and i?4 for any load P passing over the structure. 
Let the influence line for R2 be first determined. If the 
load P be placed at R2 the reaction at that point must be 
equal to P and an ordinate may be erected (Fig. 5) below 
that point on the base line ST, equal to P. It is evident 
that for a load between Ri and R2 the reaction R2 varies 



148 



C/INTILEVERS. 



[Ch. IV. 



directly as the distance from 7?i ; that is, it is equal to 

P-j- and is graphically represented by the line SU. For 

any load between R2 and the hinge at A, the variation of 
i?2 will be indicated by a continuation of the line SU to 
the point V vertically below A, since R2 still remains 

equal to P-r . 

A load P placed at the point B fails to affect the reaction 
R2, being carried entirely to Rz and 7^4. Between B and 




A, R2 varies directly as the ordinates of a straight line, 
for precisely the same reason that all influence lines are 
straight lines between panel points ; since the points W and 
V of this line have been determined, this portion of the 
influence line is represented by VW. The ordinates be- 
tween ST and SUVW therefore represent the variations 
of i?2 as any load passes over the stnicture. It is evident 
that no load between the points B and i?4 influence the 
reaction R2. 



Art. 31 REACTION INFLUENCE LINES FOR CANTILEVERS. 149 

In an exactly similar manner the influence line for R^ 
may be represented by the line TXYZ, the slope of the 
line TXY being found by erecting at Rz an ordinate of 
height P. 

The influence line for the reaction at Ri is represented 
in Fig. 6. For a load at Ri the reaction is P and is' repre- 
sented by the ordinate S'U' erected at that point. As 
the load P passes from Ri to R2, Ri varies as the ordinates 
of a straight line, the value of Ri being zero when P is at 
i?2. This rate of variation remains constant as far as the 
point .4, it being noted, however, that after passing R2 
the reaction Ri becomes negative. This is indicated by the 
diagram, where the ordinates between the base line S'T' 
and the line l^'IF' are drawn below the base line. The 
variation in the reaction 7?i for loads on AB is represented 
graphically by the line W'X', and is found in precisely the 
same manner as for R2 and R3. 

As a general conclusion, it is clear that for the greatest 
upward value of Ri, the distance S'V must be covered 
with load, and for the greatest downward reaction or uplift 
the distance V'X' must be covered. Loads on other por- 
tions of the structure do not affect the reaction at Ri. 
The influence line for R4 is also represented in Fig. 6, 
but requires no detailed explanation. 

Inspection of Figs. 5 and 6 vnll show. that the slopes of 
the pairs of lines SUV and U'V'W, and YXT and Y'Z\ 
are respectively the same. This must evidently be the 
case, for at any point in the spans represented by these 
lines the algebraic sum of the pairs of reactions, viz., of 
Ri and R2, and of Rs and i?4, must always equal P. It is 
to be noted that the distance AB, or the suspended span, 
is not included in these lengths. 



ISO 



CANTILEVERS 



[Ch. IV. 



Art. 4. — Shear Influence Lines for Cantilevers. 

'The shear influence Hnes may at once be drawn by- 
means of the reaction influence Hnes. Let the shear in 
the section MN situated between Ri and R2 (Fig. 7) be 
required, that figure representing a part of a cantilever 
structure resting on the supports Ri, R2, Rs ■ ■ ■ The 
base hne ST (Fig. 8) and the reaction influence Hne for i?i, 




UVWX, are first drawn. For any load between the hinge 
B and the section MN the shear for MN is always equal 
to the reaction Ri, but between MN and i^i the reaction 
must be diminished by the amount of the load P, as shown 
in the diagram. The resulting shear influence line for 
the section MN is then the Hne XWVYZS. It is clear 
then that for the greatest positive shear at MN the span 
R1R2 must be covered by the load extending from Ri 
to the section, while the cantilever-arm R2A and the 
suspended span AB are also covered. There should be 
no load between the section MN and R.i. For the greatest 



Art. 4-] SHEAR INFLUENCE LINES FOR CANTILEyERS 151 

negative shear, however, the load should be placed only 
between MN and R2. 

In a precisely similar way the shear influence line for 
the section M'N' , at any point in the cantilever -arm R2A, 
may be drawn. The reaction influence line for R2 (Fig. 9) 
is drawn and that portion of it included between the section 
M'N' and the hinge B represents the shear influence line 
for the section M'N' . The shaded portion of the diagram 
represents the sum of all the positive shears. It is evident 
from this diagram that positive shear only can exist at 
M'N' and for the maximum value the suspended span 
and also that portion of the cantilever span included between 
the section M'N' in question and the hinge A should be 
covered with the load. The lead en other portions of 
the structure does not afl'ect in any way the shear in the 
cantilever-arm. 

These two cases sufflce to illustrate the variations of 
shear in any cantilever structure. If the latter have 
parallel and horizontal chords, the maximum web stresses 
may at once be obtained by multiplying the maximum 
shears by the secants of the inclinations of the web members 
from a vertical. The treatment of influence areas is 
at once applicable to cantilevers, for the span lengths 
being usually large, a uniform loading may replace the 
locomotive concentrations without essential error. Loco- 
motive concentrations require the use of a criterion for the 
exact determinations of stresses. 

The division of the structure into panel lengths has 
not been treated in detail for shears, nor will it be for 
moments, for it is now well established that the effect of a 
panel on a general influence line is simply to remove all 
angles from it below the panel. 



152 



CANTILEVERS. 



[Ch. IV. 



Art. 5. — Moment Influence Lines for Cantilevers. 

Fig. 10 represents partially a cantilever structure in 
which it is desired to obtain the influence line for moments 
at the section 7l/A^ in the anchor-arm R1R2. For loads 
between Ri and R2 it is evident that the moment influence 




for b^fliua M'N' 



line for MN is exactly the same as that for a simple non- 
continuous structure. It is only necessary to construct 
ordinates equal to x and x' on the base line ST (Fig. 11) 
at the points 5 and V below the reactions Ri and R2 respec- 
tively, and connect the ends of these lines to 1' and 5. 
The influence line for that portion of the span is then the 
line SUV. 

The moment at the section MN for any load on the 
cantilever-arm R2A is represented by Rix. It is seen 
that this quantity varies at the same rate as Ri, since 
X is a constant in this expression ; R-^x may therefore be 
represented by a continuation of the line UV to W. 
Since Ri is negative for this portion of the span, the 



Art. 5 ] MOMENT INFLUENCE LINES FOR CANTILEVERS. 153 

• 

moments are also negative, as represented by the influ- 
ence line. As the load at B causes no bending at AIN 
the variation of the moment for a load passing along 
AB is represented by the line WX. The final influence 
line is therefore SUVWX, the portion SU\^ representing 
positive moments and the portion 1/VFA' negative moments. 
Positive moments may be regarded as moments causing 
compression in the upper chord members and tension in 
the lower chord. 

The influence line for moments at any section M'N' 
in the cantileA^er-arm is represented by Fig. 12. For a 
load between M'N' and A the moment varies directly 
as its distance from M'N' , and it is only necessary to erect 
below A an ordinate X' V equal to d or the distance between 
M'N' and A , and to draw the line U'V ; this line then 
represents th3 variation in the moment at M'N' , for the 
distance shown. For loads on the suspended span AB the 
variation in the moment is then represented by V'W , and 
the completed influence line is U'V'W . 

■ It is thus shown that all moments in the cantilever-arm 
are negative or such as to cause stresses of tension in the 
upper chord members, and that for their maximum values 
the entire suspended span and that portion of the cantilever- 
arm between M'N' and A must be covered with load. 
Loads on other portions of the structure in no way affect 
the moment at M'N'. 

Stress infl.uence lines for the chord members of canti- 
levers are not required, since the moment influence lines 
also suffice to furnish the stresses in those members by 
dividing by the proper lever-arms. It is only necessary 
to measure the various shaded areas shown in the figures 
and to multiply them by the proper intensity of the uniform 
load and then to divide by the lever-arm. To determine 
the scales of the various figures, it is sufficient to place a 
unit load at any convenient point and obtain the desired 



154 CANTILEyEKS. ' [Ch. IV. 

moment for that position. This determines the value or 
the scale of the ordinate at that point, and therefore the 
scale of the entire figure. 



Art. 6. — Stress Influence Lines for Members of Cantilevers. 

Fig. 13 illustrates a portion of a cantilever structure 
in which the span UqU^ is the anchor-arm, U^U^ the 
cantilever-arm, and UgUie the suspended span. Other 
portions of the structure to the right of Uie do not affect 
the stresses in any of the members shown in Fig. 13. 

Fig. 14 illustrates the influence line for stress in the. 
member U^Ly of the cantilever-arm. Passing a section 
through U(,Ut, UeL-j, and L^Lj, the centre of moments 
falls at a, projected vertically to the base line MN at C. 
As a load passes from U7 to Ug the stress in UeLy varies 
directly as the distance of that wheel load from a or C, 
for the stress in UgL^ equals the moment about that point, 
divided by the lever-arm ab, and it may be represented 
graphically by BD, a portion of the line BC. A load at 
either Ue or Uie causes no stress in UeLy. The final line, 
therefore, is ABDN. 

Fig. 15 shows the influence line for web member U2LS 
of the anchor-arm; for a load on the span U0U5 the stress 
may be represented by the influence line FGHIJ, precisely 
as for a non-continuous simple span. The line // is then 
continued to K, the vertical projection of t/g, and then 
connected to A^. The final line is the line bounding the 
shaded area shown in the figure. 

Figs. 16 and 17 show the influence lines for the members 
Lgt/g and U2UZ respectively. They require no detailed 
explanation, for they represent the variation of the bending 
moments at the points Uq and t/3 respectively. 



Art. 6. ] 



STRESS INFLUENCE LINES FOR MEMBERS. 



155 




156 CANTILEVERS. [Ch. IV. 

Fig. 18 represents a complete cantilever structure, for 
which UqUq and f/ 196^^2 5 are the suspended spans, U&U^ 
and ^16^19 the cantilever-arms, and UqUiq the anchor- 
span. 

Fig. 19 exhibits the influence line for stress in U11L12; 
the portion BCDE is found by treating the load on the 
anchor-span as if the latter were a simple non-continuous 
structure. The centre of moments falls at h and is projected 
vertically downward to F, from which point the lines BC 
and DE are drawn through points directly below the ends of 
the span U^Uie,. BC is then continued to A and DE to G, 
the ends of the cantilever-arm. These points are connected 
to 71/ and A^. The final Hne is, therefore, MABCDEFGN, 
and shows the position of loading for both maximum and 
minimum stress, and also the stress area. 

Fig. 20 is drawn in an exactly similar way for the 
member UiqLh, but in this case the centre of moments 
falls at a, within the anchor-span, and the influence line 
suffers the changes shown. 

Fig. 21 is the influence line for stress in the chord 
member U ^JJ ^^, the centre of moments for which is pro- 
jected down to 5. 

Cantilever on Towers. 

A constructive device used advantageously in cases 
where the cantilevers are supported on high iron or steel 
piers or towers is shown in Fig. 22. It leads to appreciable 
economy by the shortening of the clear cantilever opening 
as well as the anchor-arm. It consists in separating the 
feet of the two inclined posts L^Ls and L9L10 to some con- 
venient distance and omitting all bracing in the rectangle 
UgUgLsLg. No shear can then be transferred past Lg to 
the left or past Lg to the right. Since this last condition 
exists, the reaction Ri at Lq, and the positions of loading 



Art. 6. 



STRESS INFLUENCE LINES hOR MEMBERS. 



157 




158 



CANTILEVERS. 



[Ch. IV. 



for all the maximum web and chord stresses in the cantilever 
structure, as foiind from the preceding constructions, will 
hold for this case without any change whatever. It is only- 
necessary to observe carefully that the open panel U^Uq 
is to be treated as reduced to a point or entirely neglected. 
This is shown in Fig. 26, which is the influence line for 



Up U^ Uj U3 U^ UrfT U„ U, UgU^ U;ii Un U;, U13 Uu Ui5 U16 U17 Uis LI|o U 



FIG. ... ^^^PS%]i 



Fig. 23. M 



Fig. 24. M 



Fig. 25. Ml 



Fig. 26. 




lafluenceJuiB-for R 
G 



TTrnTTTmMlTTIjllIf^ L 



Influeuce line foij-Ro 



■-^^nniiiinTf]]]^^ 



Influence line foij Rj^ 
I 
I 
I 



R s-^"Ullllilllj|^^ If^ 



UqLc,. The line ST is parallel to QR, and would form a 
continuation of it if the point 5 were transferred to R. 

The only changes necessary are for the reactions R2 and 
Rz, at Ls and Lg respectively, and the determination of 
the stresses in the members UsUq and LgLg. These latter 
stresses are always equal to each other, but with opposite 
signs; they may be found by obtaining the maximum 
moment at Lg and dividing by the height of tower, UgLg. 
The reaction influence lines for Rs and R2 are shown in 
Figs. 23 and 24. 

In the case" of Fig. 23, the ordinate below Ug is equal 
to the load itself, or P, and diminishes as a straight line BA 
to Us- R3 remains equal to P for the load between U9 



Aet. 7.] 



SUSPENDED CANTILEVERS. 



159 



and the end of the cantilever at L'13, but diminishes as 
shown by CN from that point to U20, being always equal 
to the reaction of the simple suspended span at Uiz. 

Fig. 24 exhibits the variation of the reaction at R2. 
It is equal to P when the latter is at Lg, but it equals zero 
for the load at Uq, Uq, and U20. The slope of the hne FG 
is a continuation of the slope of AIE, i.e., FG is parallel 
to AlE. The final influence area is that shown shaded. It 
is seen that no uplift ever occurs at R2. 

The variation of Ri as the loadP passes over the struc- 
ture is shown by Fig. 25, and it differs in no way from 
that found without the open panel ^8^9- The line HI 
is parallel to JK. 

As a check on the construction of these three figures, 
it is only necessary to note that at any one point the algebraic 
sum of the ordinates for Ri, R2, and Rs must equal P, 
the load. The diagrams fulfil this condition. 

Art, 7. — Suspended Cantilevers. 

The structure shown in Fig. 27 is a true cantilever, 
although it possesses some of the lines of a suspension 
bridge, and the term suspended cantilever may be applied 




Fig. z-j. 



to it. Some general deductions as to its statical condition, 
and the application of the method of influence lines to 
determine the maximum stresses, will be found in the 
following article: 



159' CANTILEVERS. [Chap. IV. 



Statical Condition of the Structure. 

Fig. 2 7 shows a coplanar frame work resting on the four 
points of support A, B, C, and D, and any one of them may 
be considered fixed while all the others are supported on 
nests of rollers moving on horizontal beds. These conditions 
of support eliminate temperature stresses. The structure 
is then virtually a stiffened suspension bridge with vertical 
anchorages, the cable being AKEFGHLD, while the road 
bed lies along the lower chord A BCD. 

If the structure be inverted and supported at the points 
AKLD of which only one is, as before, fixed, the framework 
becomes a continuous bridge of three spans, not subjected 
to temperature stresses ; it may be considered a stiffened 
arch with vertical reactions. If the structure is supported 
at K and L as fixed points it becomes a two -hinged arch 
subjected to temperature stresses. 

In all these cases, even where the reactions are vertical, 
there are four such reactions to be found ; therefore, Fig. 2 7 
is a statically indeterminate framework ; that is, the three 
equations of condition of coplanar statics are insufficient for 
the determination of the stresses in the members. There 
are required, in addition, other equations of condition, 
depending either on the elasticity of the material or the 
work performed in the structure by the displacement of its 
members. 

The structure, shown in Fig. 27 may, however, be made 
statically determinate by the removal of the members 
EF and GH, as shown in Fig. 28, and by assuming the two 
panel points M and N to be frictionless pins. The structure 
is then a suspension cantilever in which the spans AB and 
CD are the anchor arms, the spans BM and NC the canti- 
lever arms and MN is the suspended span., No difficulty 
will be experienced in determining for this class of structure 



Art. 7.] 



SUSPENDED CANTILEVERS. 



159* 




Fig. :8. 




IS9C CANTILEVERS. [Chap. IV. 

the stresses due to dead load. The method of influence 
lines is, however, particularly applicable for determining the 
positions of the moving load causing the maximum stresses, 
as well as for determining the maximum stresses them- 
selves. 

Fig. 29 represents a suspension cantilever, the span 
/i = LoLq being the anchor arm, the span 12==LqL2 being the 
cantilever arm and the span /3 =L:)Lii being the suspended 
span. The entire structure is taken to be symmetrical 
about the center of the panel LnLi2. The distances 
measured between the lower panel points L and the upper 
panel points U are designated by h with a proper subscript ; 
for instance, U^^L^ is h^ and UqLq is hf,. 

It will be assumed that the points Lo and L23 are 
anchored and capable of carrying negative or -downward 
reactions; the points Lq and L^ are supported on nests of 
rollers moving on horizontal beds. For the static deter- 
mination of the stresses and for the elimination of tem- 
perature stresses it also becomes necessary that one of the 
points of support of the suspended span be so carried that 
it may move horizontally over the end of the supporting 
cantilever arm ; the point L14 of the suspended span may, 
for instance, be placed in a slotted hole. 



Art. 8. — Fixed Load Stresses. 

Fig. 30 shows a portion of the structure of Fig. 29 for 
which it is desired to obtain the stresses due to fixed load. 
For case of illustration, the dead loads W-iW'^-i . . . PF9 are 
all assumed to be supported at the lower panel points only. 
Their actual distribution between upper and lower chords 
would add, however, no difficulty to the work. 

The reaction i?o at Lo may be obtained at once by taking 
the moments of all external forces about Le and placing 



Art. 8.] 



FIXED LOAD STRESSES. 



159'/ 



them equal to zero. It is to be observed that the weight 
Wq includes one half the weight of the suspended span. 
The equation involving the reaction Rq as the unknown 
quantity will then have the form 

Rlx=±lWx 

where x represents the distance of any panel weight from 
Le and is to be taken plus or minus as it is measured to the 
right or left of Lg. 

The usual force polygon may then be constructed, 
taking at first the unknown forces LqUi and L Li acting 
at panel point Lq and balancing the reaction i?j. In order 
to simplify the terminology, the members will for the 



•-»* 




Li L3| L,. L 

W, ■'Wz '''W3 -fWi 
—l-i 






■-8 |L9 Li 



Fig. 30. 



present also be designated by the numbers placed upon 
them. Having found, therefore, the stresses in the members 
I and 2, the stresses in the members 3 to 10 inclusive, taken 
in numerical order, may then be determined. At panel 
point Uz there are, however, three unknown concurrent 
forces, and one of them must be known before the others 
can be found by means of the force polygon. 

If a section ST be passed through the members 11, 12 
and 13, the stress in the member 1 1 may be found by taking 
moments of all the external forces situated to the right of 
the section (namely, Wy, W^ and W^) about Lo as the center 



i59e CANTILEVERS. [Chap. IV. 

of moments. The stress in member ii, multiplied by its 
lever arm about Lq is then equal to the sum of the moments 
of these external forces. The stress in member ii having 
thus been obtained, the stresses in members 14 and 15 may 
at once be determined, for they are the only two unknown 
forces acting at Uq. Panel point U5 may then be treated 
in turn, and the stresses in the members 16 and 17 found. 
In precisely the same way there may then also be deter- 
mined the stresses in the members 18 and 19. 

One of the three unknown stresses at panel point U3 has 
thus been found, and the determination of the stresses in 
members 20 and 21, or in any other members of the 
structure, presents no further difficulty. 

Art. 9. Live Load Stresses. 

In treating the live load stresses in the various spans of 
Fig. 29 it is first to be noted that the stresses in the sus- 
pended span I3 are precisely the same as if that structure 
were a simple non-continuous span; it requires, therefore, 
no further consideration. 

The anchor span /i is also a simple non-continuous span 
for any loads placed on that span only, the members UzUi, 
U4U5, Uid, etc., not acting. Loads on the spans h and 
I3, however, cause stresses in the members of this span and 
these stresses must be investigated. Since they are caused 
by the stresses in the members U3U4, U4U5, and UsUq, and 
the suspender rods U^Ci and U5C5, these stresses must first 
be found. The horizontal components of all the upper 
chord members, such as UiU^, must necessarily be equal, 
since the only other members which are attached to them 
are vertical and carry no horizontal components. An in- 
fluence line for the variation of this horizontal component 
as a load passes between the points Lq and L^ will there- 
fore be required. , 



Art. 9] LIVE LOAD STRESSES. 159/ 

Let there be placed a load P at Lg, the end of the can- 
tilever arm I2, and let a section be passed through the 
members UqUj, LqCj and LqLt. The horizontal component 
of the stress in UeUj may then be found by taking its 
moment and those of the external forces about Lq. The 
vertical component of UqUj, disappears in the moment 
equation since its lever arm about the center of moments is 
zero. If H represents the horizontal component, the follow- 
ing equation will result : 

H-h, = P.l2, (i) 

or 

P- U 

«=Ar (^' 

Equation 2 shows that the value of H varies directly as 
the distance of the load P from Lq ; this variation may 
therefore be indicated by a straight line for the span I2. 
The influence line for H may therefore be drawn as follows. 

(Fig- 31): 

At the point D, situated on the base line AC directly 
below Lg the ordinate BD is erected with a value H =P ■ h/h^,, 
and the point B is then connected to ^ by a straight line. 

Any load on the span h is divided by the principle of 
the lever into two component parts, acting at L9 and L14 ; 
that part at L14 causes no stress whatever in that portion 
of the structure to the left of Lg. That part of the load 
carried to L9 varies directly as the distance of that load 
from Lq and its effect on the influence line H may be repre- 
sented by a straight line between Lg and L14, with a zero 
value at L14. In Fig. 31 the influence line is therefore com- 
pleted by connecting the point B to the point C. It is to 
be noted that this distribution of the load on the span h 
causes every influence line between Zg and L^ to be a 
straight line. Hereafter, therefore, if the value of the 



■159.?- 



CANTILEVERS. 



[Chap. IV, 



ordinate below Lg is known, its end may at once be con- 
laected to a point on the base line directly below L14. 
; It is seen that every load on the spans I2 and /s increases 
,;tlae value of H ; the maximum value of H is therefore found 
by covering the spans h and l^ completely ; if the load be 
.taken as a uniform load (which it is proper to do for long 
spans) the area of the triangle ABC multiplied by the 
intensity of the uniform load will furnish the maximun" 
value of H. 

Let the following data apply to this bridge : 

f/iZ-i = 3o feet = /zi, 

£72^2 = 38 " =^H, 

UsLs^^o " =UQLc) = UioLio=UnLii = h3 = h9, etc., 

6^4^^4=45 " =UsL8 = hi=h8, 

(74X4=38 " —CsLs, 

£/5l'5=55 " =U7L7 = h5=h7, 

CbL-^=2,o " =C7L7, 

All panel lengths are taken equal at 30 feet each ; the 
span /i is therefore 180 feet, the span I2 90 feet, and the 
span ^3 1 50 feet. 

The value of H for a load of unity at Lg is therefore 

'— = 1.2 and the ordinate BD is drawn with that height 
75 ■ 

The area of the triangle ABC is then — = 144, and 

if the intensity of the loading be taken at 2000 pounds per 
linear foot of truss, the maximum value of H becomes 
288,000 pounds. The maximum stresses in the vertical 
suspender rods and upper chord members UsUi, U4U5, etc., 
will then be found by means of the simple force diagram 
^hown in Fig. 32. For clearness of illustration the right 
hand tower UnL^ instead of the left-hand tower is treated. 
The horizontal force H is first laid down to proper scale 



Art. 9.] • LIVE LOAD STRESSES. 159// 

and from one end of it are drawn lines parallel to UnU is 
and UicJhi until these lines intersect the vertical gh erected 
at the left end of H. These lines, when measured to scale, 
furnish the maximum stresses in those members. The stress 
in the tower column UiyLn, is found in the same force tri- 
angle ; its value is the distance gh or the ^'ertical intercepted 
by the lines just drawn. 

The maximum stresses in the suspender rods and upper 
chord members are found in the same figure by drawing 
lines respectively parallel to L/j^L^'ig, U1QU20, U iJJ i^, etc. 
The stresses found for these members are clearly indicated 
in the figure. They are found from the simple principle of 
the .force polygon as applied to the equilibrium of concur- 
rent forces for at any point, such as Uks, there are to be 
determined only two unknown forces Ui^Uiq and UiqCiq. 

Art. 10. — Influence Lines for the Chord Members of the 
Anchor Span, ZoZ-e- 

The influence line for the member C^Lq of the span li 
will first be drawn. It has already been shown that, for 
loads on the span /i, the members of the anchor span are to 
be treated as if that span were a simple non-continuous 
span. For that condition, then, the stress in the member 
C^La is found by passing a section through the members 
b\U(i, C^Lq, and L^Lq. Since the member U^Uq is not 
stressed, the center of moments may be taken at L5. 

Since the stress in C^L^ is equal to the moment at L^ 
divided by a lever arm, its variation may be represented by 
the variation in the bending moment at L5, and Fig. 33 
represents the influence line for the moment at L5. For the 
span /i this line is EFJ ; the value of the ordinate FG is 
found by placing a unit load (i pound) at L5; the reaction 
at Lq is then Rq = ^ and its moment about L5 i"s J- X 150 • 
21:5 foot-pounds or FG. 



159^ CANTILEVERS. [Chap IV, 

When loads are placed on the pans I2 and Iz the member 
U^U^i is stressed ; but in taking moments about L5 for the 
stress in the member C^Lq it will only be necessary to con- 
sider the horizontal component of the stress in U^Uq, for 
the moment of its vertical component disappears, its lever 
arm being zero. 

Let there be placed a load P at Lg and let it be required 
to find the moment at L5 for that position of the load. The 
reaction at Lq is upward and is equal to 

R^ = ^ (3) 

'-1 

If H represents the horizontal component of the stress in 
any upper chord member, the bending moment M5 at L5 is 
then given by the following equation : 

M,^Ro-x,~H.h, (4) 

Substituting the values of Rq and H from Eqs. (3) and 

^-K'lf-^O w 

Substituting the numerical values of these quantities, 
and assuming P to be a unit load, 

/go X I so 90 X sS \ 
M5 =( J ^^^^ - ^g^^ j = 9 foot-pounds. 

The ordinate KN below Lg is drawn downwards with 
this value, for it causes a tensile stress in C^Lq. 

It is to be noted that the value of M5 as given by Eq. (5 ) 
varies directly as I2, or the distance of the load from Lq. 
The influence line (Fig. 33) for the spans I2 and I3 may 
therefore be completed by drawing straight lines from N to 



Art. 10.] INFLUENCE LINES FOR CHORD MEMBERS. 159/ 

J and P. The shaded areas show the portions of the struc- 
ture to be covered by loading, to cause the maximum 
stresses of opposite kinds ; it is to be remembered, however, 
that these areas are moment areas, and are to be divided 
by the lever arm of C^Lq in order to furnish stresses. 

The value of the maximum moment, causing compression 
in CqLq is 

T-T-r 180X25x2000 
area hrJ X 2000 = = 4,800,000 loot-pounds, 

while that causing tension is 

TATD 9 X 240 X 2000 . r . . 

area J JMFx 2000 = = 2,160,000 toot-pounds. 

The influence line lor the member U^Ci is shown in the 
same figure as EF'J for the span /i and as JN'P for the 
spans I2 and I3. For the latter spans, the stress is compres- 
sive just as for the span /i, as the evaluation of Eq. (5) shows. 
Moments are taken about the panel point L3, and there is 
found, for a load of i pound at Lg: 

,^ 90 X 90 90 X 40 . , 

M3 = ^g^ — — = - 3 foot-pounds. 



The ordinate G'F' is found by placing the unit load at 
is; the reaction Rq becomes r^-, and the moment J X 90 f= 45 
foot-pounds. 

The influence line for a chord member, such as U1U2 is 
found in precisely the same manner by means of the same 
equations; in Eq. (5) however the term involving H dis- 
appears, and there is to be used 

M. = P^ (5a) 



159-^ CANTILEVERS. [Chap. IV. 



Art. II. — Influence Lines for the Chord Members of the 
Cantilever Arm L^L^. 

The influence line for the member CyCg will be drawn ; 
it is at once evident that loads on the span h cause no 
stress in any of the members of the spans h or l^, and that 
span need not be considered. 

The stress in C-C^ is found by passing a section through 
C/yC/g- CjC^, CyLs, and LjL^, and then taking moments 
about Zg. The stress in U-jUg may be found, for any posi- 
tion of the load ; its horizontal conponent need only be con- 
sidered, for the moment of the vertical component dis- 
appears as its lever arm is zero. 

If a load P be placed at Lg, and if Mg represent the 
moment of the stress in CyCg about Lg ; 



But 



therefore 



Mg =H-hs, (6) 

^_ Pil2 -a) . 






The stress in C7C8 will result at once from dividing Mg by 
the normal distance from Zg to that member. 

For any position of the load between Lq and Lg, Eq. (7) 
shows that Mg varies directly as the first power of the 
variable distance {I— a); this variation may therefore be 
represented by the straight line SV in Fig. 34, the ordinate 
VU directly below Lg having the value furnished by Eq. (7) . 



Art. 12.] INFLUENCE LINE FOR IVEB MEMBERS. 159/ 

If a load be placed at Lg, the moment becomes 

Ms=Hhs-Pa (8) 

The value of the ordinate XW of the influence line erected 
at Lg is therefore found by continuing the line SV to inter- 
sect the vertical dropped from Lg and subtracting from it 
P a. 

The influence line is then finally completed by connect- 
ing X and r by a straight line. 

Substituting the numerical quantities in Eq. (7) and 
assuming P to be i pound, there is found, 

Ms = ^90- 30)45 _ ^g foot-pounds = UV. 

The value of XF is P a =30 foot-pounds. 

The shaded area of Fig. 34 when multiplied by the 
intensity of the uniform load, furnishes the maximum 
bending moment at Lg. When divided by the lever arm 
there will be obtained the maximum stress, which is com- 
pressive. 



Art. 12. — Influence Line for Web Members of the Anchor 

Span LoLg- 

The influence line for the member C4L4 will be drawn ; 
its stress is found by passing a section through UiU5, CaP-^, 
C4L4 and L3L4, and dividing the bending moment found at 
the intersection d of C5C4 and L3L4, distant b from Lq, by 
the lever arm of C4L4. It will then only be necessary to 
draw the influence line for moment at that point; the 
stress in C4L4 may be found directly from that moment. 



I59W CANTILEVERS. [Chap. IV. 

Load on Span li. 

The influence line for the load at any point of the span 
/i, is found just as if that span were a simple truss ; lines are 
drawn (Fig. 35) from any point K\ found at any point 
directly below the center of moments, to points on the base 
line A' and E' , directly below the ends of the simple span. 
The lines A'B' and D'E' are then portions of the influence 
line. In the panel LsL^, the influence line is also a straight 
line, and the line is finally completed by connecting B' and 
D' by a straight line. 

If a load P be placed at L4, the reaction at Lq becomes 
in the present case Ro =--- ^P ; and the moment of the external 
forces situated to the left of the section passed, is 



M.= ^('.0) = p(i«^5)^87.5P. 



If P be taken equal to i pound, the value of the ordinate 
D'H' (Fig. 35) is therefore 87.5 foot-pounds. This value 
determines the scale of this portion of the influence line. , 

Load on Spans I2 and I3. 

For a load placed on the spans I2 or I3, it will be neces- 
sary to obtain the stress in the member U^Us, for that 
stress appears as an unknown quantity in the equation of 
moments about the center of moments d. 

It will be more convenient, however, to treat the hori- 
zontal and vertical components, H and V, of that stress. 

Placing a load P at Lg, the moment M^ at d, then be- 
comes : 

M^ = Ro{l, + b)-Hh4~V{b + h-X4), ... (9) 
but 

Ro-Pt, H=Pr, and V = Htana, 



Art. 12.] INFLUENCE LINE FOR WEB MEMBERS. I59« 

where a is the angle between the member UaU^ and a 
horizontal. Since tan a = — ^— — , where ^ = panel length, 



y^p. h {h5-hi) 



Substituting these quantities in Eq. (9), 



Ma^P 






Inspection of Eq. (10) shows that, for a load placed between 
Le and Lg, 71/^ varies directly as I2 or the distance of the 
load from Le ; this variation may therefore be represented 
by a sti-aight line E'F' in Fig. 35, the ordinate /'F' being 
given by Eq. (10). If P equals i pound, that equation 
becomes, for the example treated : 

^. = ^(180 + 82.5)-^- 45-(82.5 + i8o-i2o)^Q 

= 20^ foot-pounds. 

The stress corresponding to this moment is compressive. 
The influence line is completed by connecting the points F' 
and G' by a straight line. 

The final influence line (Fig. 35) shows that the greatest 
tension in CiL^ occurs when the distance C'E' is covered by 
the uniform load ; the greatest compression is found when 
spans I2 and Z3, and the portion A'C of span li are covered. 

It is to be remembered that the shaded areas shown in 
Fig. 35 are moment areas, and must be divided by a lever 
arm to furnish stresses. 



159^ 



CANT/LEVERS. 



[Chap IV. 



The influence line for any web member, such as U2L2, 
not lying below the cable members UsUi, etc., is also found 
by the use of Eq. (9), precisely as in the case of C4L4; but 
the terms in that equation which involve H and V dis- 
appear. 



Art. 13. Influence Line for Web Members of the Cantilever 

Arm LqLq. 

The influence line for the members CgLg, of the canti- 
lever arm is shown in Fig. 37. The necessary parts of the 




Fig. 37. 



structure for determining that line are shown in Fig. 36. 
Loads placed on the anchor span /i cause no stresses in this 
member. 

The stress in CgLg is found by passing the section ST 
shown, and dividing the moment of the proper external 



Akt. 14.] INFLUENCE LINE FOR IVEB MEMBERS. \S9p 

forces taken about ^ as a center, by the lever arm of C^L^. 
A.S in previous cases, therefore, the moment influence Hne 
for the moment Mg of the external forces taken about e will 
be drawn, since the stress in C^L^ may be found directly 
from it. The horizontal and vertical components of stress 
in U^Us are denoted by H and V respectively, while m and 
Xs are the distances shown in the figure. If a load P, of 
one pound, be placed at Lg, the equation of moments about 
e, treating U^Us as the external force, will be 

M,='H-h^^V{m + x^) (11) 

But 

H = —J and V = H tan a, 

if a represents the angle between UjUg and the horizontal. 
Therefore, 

n/r P -xs-hs P ■ xsjm + %) 

M e = J 1 j tana. . , (12) 



Eq. (12) shows that Me varies directly as x^; since this 
term appears in the first degree, the equation may be repre- 
sented by the straight line BA in Fig. 37, the ordinate BC 
having the value furnished by Eq. (12). For the structure 
under consideration, Eq. (12) becomes 



60 X 45 , 6 0(82.5+ 60) I ^ . . , 

Me = — z 1- ~ = 74 foot-pounds. 

75 75 3 



The influence line passes through A, a point on the base 
line, for a load at that point causes no stress in CsLg. 



159(7 



CANTILEVERS. 



[Chap. IV. 



If a load P of I pound be placed at £9, the equation for 
Me becomes 



M„ = ; 1 J-. — ~ tan a — P{l2 + m) . (13) 



h. 



ha 



Inspection of Eq. (13) shows that the first two terms of 
the right-hand member represent a continuation of the line 
expressed by Eq. (12). The third term shows that there 
must be subtracted from each ordinate of that line between 
Ls and Lg, the value of P multiplied by its distance from e. 

In Fig. 37 there must be subtracted from the ordinate 
EH the moment P(9o + 82.5) = i72.5 foot-pounds. 

The final influence line is therefore ABDG; the area 
FDG indicates a tensile stress, and the area ABF a com- 
pressive stress. 



Art. 14. — Suspension Cantilever with Two Points of Support 
at Intermediate Pier. 

Fig. 38 shows the suspension cantilever, treated in the 
preceding articles, but in which the post U&L^ has been 



Us Ug' 




=1 f'-o Lj Lg IL9 Liu Lii L, 
-^-l- ' !-., *k ! 



Fig. .^8. 



replaced by two columns, connected at top and bottom by 
horizontal members UqU^ and LqLq, but otherwise with ro 



Art. t.;.1 suspension CANTILEVER. 159;- 

bracing between them. No shear can then be transferred 
past Lq' to the left or past Lq to the right. No change 
occurs, thefore, in any of the preceding constructions in 
determining Rq or any of the stresses in the structure ; it is 
necessary to observe that the open panel LqL^' is to be 
treated as a point, or entirely neglected. Changes occur, 
however, in determining the reactions Rq and Rq at Lq and 
Lq respectively; the stresses in the members UqUq and 
LqLq must also be found. These conditions are precisely 
the same as those for the ordinary cantilever structure,* 
supported on towers. 

* See Page 158. 



CHAPTER V. 

DEFORMATION OF TRUSSES. 

The members of a framed structure when stressed 
suffer strains or deformations and these strains cause dis- 
placements at all points in the structure. These displace- 
ments may be found by graphic methods if the strains 
in the members are known. The deformation of any 
member may be expressed by the following equation: 

SI 
^^ = ^ + < (i) 

in which Jl is the change in length of any member, expressed 

in inches, being positive for tensile stress and negative for 

compressive stress, S the total stress in the member, 

expressed in pounds, A the area of cross-section of the 

member in square inches, / the length of the member 

in inches, E the coefficient of elasticity in pounds per 

square inch, e the thermal linear coefficient of expansion 

of the material in the member, per degree Fahrenheit, and 

t the change in temperature in degrees Fahrenheit. In 

order to find the displacement of the points of a structure 

it is necessary to consider the simultaneous strains or 

stresses caused by one position of loading and not the 

maximum strains or stresses, since the latter occur for 

different positions of the loading. 

1 60 



Art. 1.] 



THE DISPLACEMENT DIAGRAM. 



i6l 



Art. I. — The Displacement or Williot Diagram. 

A simple statically determinate truss is a structure 
formed in general by adding to some triangle as a base 
element two additional members for each panel point in 
the frame. The problem of determining the displace- 
ment of any point in a simple statically determinate frame- 
work after the latter has been subjected to stress may there- 
fore be treated in the most general way by obtaining the 
displacement of a panel point such as h (Fig. i), which 



-Ac 6 




Fig. I. 



is connected to the member ac by the bars ah and he. The 
member ac is also a member of the complete framework, 
in which the points a and c are assumed to have already 
received the known displacements aa' and cc' shown in 
the figure. By the application of stress the members ah 
and ch are assumed to receive deformations of + Aah and 
— Ach, the positive sign representing an elongation. 

Considering then the triangle ahc as still unstressed, 
the points a and c of the bar ac will first suffer displace- 
ment to the positions a' and c' . This motion displaces 



i62 DEFORMATION OF TRUSSES. [Ch. V. 

the point h of the bar ah to 61, bbi being equal and. 
parallel to aa' ; and it also displaces the point b of the 
bar be to 62, &&2 being equal and parallel to cc' . The 
member ab is increased in length, however, by an amount 
+ Jab, and this increase is represented by drawing &163 
equal to + Jab and in the direction of a' to bi. cb, however, 
is shortened by an amount — Jcb and the point 64 is found 
by laying off on c'b2, in the direction of 62 to c\ the distance 
62^4 = ^cb. As the point b cannot occupy the two positions 
63 and 64 simultaneously, its final position must be at 
the intersection of the members a'63 and c'64, and this, 
intersection may be obtained by rotating those members 
about a' and c' as centres respectively. It will, however, be 
sufficiently accurate for the small deformations generally 
found to replace these arcs of circles by straight lines drawn 
at right angles to the radii at their extremities. In this 
case, these lines are 636' and b^b', intersecting at b'. The 
displacement of the original point b is then .fully shown, 
both in magnitude and direction, by the heavy line bb'. 

.The preceding operation is more conveniently applied 
in a diagram entirely separate from the truss figure, and 
the displacement of the point b is again shown in Fig. 2, 
which is known as a displacement or Willi ot diagram. In 
that figure each displacement is referred to some fixed 
pcint known as a pole and represented by the letter 0. 
This pole represents also the zero displacement of a point 
of the framework, taken originally as immovable or fixed. 
The displacement of any other panel point is measured by 
its displacement from the pole 0. 

In Fig. 2, therefore, Oc' and Oa' are drawn through O 
respectively parallel and equal to cc' and aa' to represent 
the displacemxents of the points c and a. At c', Jcb is then 
drawn parallel to be and m the proper direction to repre- 
sent the shortening of the member be; and similarly. 
Jab is drawn parallel to ab, and in the direction a to b, to 



Art. I.] THE DISPLACEMENT DIAGRAM. 163 

represent the lengthening of that member. At the ends 
of these Hnes perpendiculars are erected and their inter- 
section at h' shows the displacement Oh' of the point h. In 
Fig. 2 the line Oh' must be the same in direction and magni- 
tude as bh' of Fig. i. 

This method is easily applied to the case of a truss- 
crane, shown in Fig. 3, for which it is required to find the 
displacement of the peak e, the point of support a being 
fixed in position. It is also assumed that the member ah 
suffers no change in direction. The crane is supporting 
only a load W at the peak. 

In displacement work the following notation will in' 
general be employed. On the truss diagram the panel 
points will always be designated by small letters, a, b, c, 
etc., while the members themselves will always be repre- 
sented by numerals i, 2, 3, etc. The deformations ii, 
i2, ^3, will, however, each be marked on the displacement 
diagram itself only by the number of the member for 
which it represents the deformation. For instance, the 
member ah in Fig. 3 is known as member i in that figure, 
and its deformation in Fig. 5 is also represented by i. 
The displaced position of any point will be represented 
by the primed letter of the point, such as a', b', c' . The dis- 
placement of such points from the pole will represent 
their actual displacement. 

Before obtaining the displacements of the points of the 
structure, it is necessary to determine the strains in the mem- 
bers. For this purpose the stress diagram shown in Fig. 4 
must first be drawn. Table I shows in detail the length of the 
members. The intensities of stress existing in those mem- 
bers are obtained by dividing the total stresses scaled from 
Fig. 4 by the respective areas of cross-section, and the last,, 
column in the table shows the positive and negative strains. 
The coefficient of elasticity has been assumed at 30,000,000 
pounds per square inch. It is the usual practice in dis- 



164 



DEFORMATION Oh TRUSSES. 

Table I. 



[Ch. V. 



Member. 


Length in Inches. 


Intensity of Stress 
in Lbs. per Sq. In. 


Deformation in 
Inches. 


ah=i 


120 


— 5000 


— .02 


ar=2 


144 


+ 7000 


+ .034 


6c = 3 


144 


-5000 


— .024 


co = 4 


120 


+ 7000 


+ .028 


M=s 


180 


— 5000 


-■°3 


d.e = 6 


96 


+ 7000 


+ .022 


6e = 7 


220 


— 5000 


-•037 



£=30,000,000 pounds per square inch 

placement work to emplo}'' the gross sections of tension 
members in place of the net sections, that is, to make no 
allowance for rivet-holes, and then to reduce slightly the 




A'i 





Scalfi In Inches 



Scale in Ponnda 



^~:^W<X, 



\ "^ 

\ 

s \ 



A 



Fig. 4- 



'6e' 



Fig. s. 



value of the coefficient of elasticity to a value of perhaps 
28,000,000 or 29,000,000 pounds per square inch. 

It has already been assumed in this problem that the 
member ah suffers no rotation in position but remains hori- 



Akt. I.] THE DISPLACEMENT DIAGRAM. 165 

zontal and that the point a remains fixed in position. The 
point = a' is therefore chosen as the pole in Fig. 5. The dis- 
placement of b with relation to that pole is found by draw- 
ing I, Fig. 5, parallel to i, Fig. 3, and equal to —.02 inch. 
The member i is in compression, and therefore the dis- 
placement of h with regard to a is in the direction of 5 to a, 
as shown in Fig. 5. If a6 were in tension, h' would be dis- 
placed to the right of the pole 0. The displacement of 
the point c may now be found by drawing lines parallel to 
2 and 3, equal to the deformations in those members and 
through the points a' and h' . The intersection of the 
perpendiculars erected at the ends of 2 and 3 will then 
determine the point c' . 

The displacement of the point d is next obtained by 
drawing through the points b' and c' lines parallel to 4 
and 5, representing in magnitude and direction the de- 
formations in those members. The intersection of the 
perpendiculars to those lines, erected at their extremities 
furnishes d' . The line connecting with d' then represents 
the displacement of d with respect to 0. In precisely the 
same way the displacement of the point e is found by 
drawing lines parallel to 6 and 7 through d' and 6' and 
erecting perpendiculars at their extremities. The dis- 
placement of e with regard to a is therefore represented 
in Fig. 5 by the line Oe' . The displacement of any other 
point with regard to a is found by connecting the pole 
with any other point. In the problem, Oe' measures 
.24 inches. 

The problem just solved is more simple than the usual 
cases of displacement found in bridge-trusses, for in those 
structures no member such as ab remains fixed in position. 
In simple bridge -trusses one end of the structure is usually 
fastened to a pin connection about which the entire truss 
may rotate, and the other pin end usually rests on rollers 
which permit that end panel point to move in one fixed 



•I 66 DEFORMATION OF TRUSSES. [Cii. V. 

.-line only In the case of bridge-trusses, therefore, it is 
. usual in the first place to consider any member as fixed in 
direction and one end of it fixed in position, then to deter- 
mine the displacements of all the points in the structure 
. with regard to that member alone. The structure, 
_ deformed and considered rigid in that condition, is then 
rotated about the end pin in such manner as to fulfil the 
.condition that the other end moves in the proper line. 
.;Two separate constructions are therefore necessary for the 
I case of bridge-trusses. The first construction is exactly 
■ similar to that already explained. The second construction 
.involves the representation of the displacements of a rigid 
figure when rotated about some fixed point. 

Art. 2. — Rotation of a Rigid Figure about a Point. 

If a rigid figure such as the shaded triangle abc of Fig. 6 
is rotated slightly, its motion may be expressed as one of 
rotation about an instantaneous centre /. If the angle of 
rotation for the triangle ahc is a, the figure then takes the 
position a'h'c' . The lines aa' , bb', cc' represent the dis- 
. placements in position of the points of the triangle ; they 
are respectivel}^ perpendicular to the lines la, lb, and Ic. 
Fig. 7 represents a displacement diagram cf a kind similar 
to that previously explained, being chosen as a pole 
from which all displacements are to be measured. Let 
.there be drawn as rays em^anating from this pole the dis- 
tances a"0, b"0, and c"0 respectively equal and parallel 
to aa', bb', and cc', these distances representing the changes 
in position of the comers of the rigid frame. Connecting 
the points a", b", and c" by the lines a"b", b"c", and c"a" 
it will then be found that the figure a"b"c" is similar to 
the original frame abc, but turned exactly at right angles 
to it, for a"0 is perpendicular to al; b"0 to bl, and c"0 
to cl ; and a"0 :al:: b"0 :bl:: c"0 : cl. 



Art. 2.] ROTATION OF A RIGID FIGURE ABOUT A POINT. 167 

It will be evident in examining Fig. 7 that if the points 
a" and c" had been known, the position of the point b" 
could have been determined by drawing upon the line 
a"c" a figure, similar, to the rigid figure ahc. ' In the general 
problems of trusses, if the rotation of any two points is 
known, the rotation of all others may be found. It will 
be found in statically determinate trusses that the rotation 
of two points is always known. , In general, one point 





such as a will be the centre of rotation, while the amount 
of the displacement of some other comer such as c, when 
subjected to the conditions of the problem, will furnish 
the rotation of the second point. 

If a point suffers two displacements, both of which are 
measured from the same pole 0, the final displacement of 
that point, with regard to 0, may be found by a triangle 
of displacements precisely similar to that of the triangle 
of forces. In Fig. 8 let a^'O represent the displacement 
of the point a of any figure found by the rotation of the 
structure about some point, and let Oa' represent the 



1 68 DEFORMATION OF TRUSSES. [Ch. V. 

displacement of the same point when the figure is deformed 
by stress. The final displacement of a is then represented 
by a"a' measured in the direction as read, and found by 
closing the triangle of displacements a"Oa' . It is to be 




Fig. 8. 

noted that the direction of the resultant in the triangle is 
opposite to that of the two component displacements. 

In future notation the displacement of any point caused 
by stress will always be represented by the primed letter, 
such as a' . The displacement caused by rotation will 
always be represented by the second prime, such as a" . 
The final displacement must then always be read in the 
direction of a" to a'. 



Art.! 3. — Deformations of a Bridge-truss. 

The preceding analysis may now be applied to a bridge- 
truss in order to find the displacements of its panel points 
when it is subject to loading. A steel railroad -truss hav- 
ing eight panels of 30 feet each with a depth of truss at the 
centre of 40 feet, as shown in Fig. 9, will be taken. The 
other truss dimensions are shown in the figure. The 
deformations due to both dead and live load will be deter- 
mined. The loading will be considered uniform, that 
being sufficiently accurate for the purpose. The moving 
train will be taken as covering the entire span. The dead 
loads or own weight are taken at 400 pounds per linear 
foot of span for the rails and other pieces that constitute 



Art. 3.] 



DEFOHM/tTlONS OF A BRIDGE-TRUSS. 



[69 




17° DEFORMATION OF TRUSSES. . [Ch. V. 

the track; at 400 pounds per linear foot for the steel 
floor beams and stringers, and 1600 pounds per linear 
foot for the weight of trusses and bracing. The moving 
train load will be taken at 4000 pounds per linear foot. 
This will make the panel loads for each truss as follows : 

Lower-chord dead load, 30 X 800 = 24,000 lbs. per panel 
Lower-chord rnoving load, 30X2000=60,000 " " " 



Total load on lower chord =84,000 " " " 

Upper-chord dead load, 30 X 400 = 12,000 " " " 

The structure is a "through" bridge, hence all moving 
loads rest on the lower chord. The stresses in the trussed 
members due to the combined uniform dead and moving 
load are easily found by the graphical method, one diagram 
only being needed. It is not necessary to show this 
diagram. 

Table I shows, then, the total stresses, the intensities of 
stresses, the lengths /, and the deformations Al in the mem- 
bers. The displacement diagram (Fig. 10), may then be 
drawn in a manner similar to that for the trussed crane, 
assuming me as fixed in direction and one point of it, m, 
fixed in position. It is usually preferable to take as the 
member fixed in position a bar which it is known in advance 
suffers little displacement. This avoids an unwieldy 
diagram and explains the choice in the present instance. 
The point m' (Fig. loj represents, therefore, the pole, and 
the displacement of e with respect to m is shown as m'e' . 
The displacement of /' may then be found by drawing 
lines parallel to 12 and 22 through the points ni' and e' 
respectively, equal to the deformations in those members 
and erecting perpendiculars at their ends. The intersec- 
tion of these perpendiculars gives the displacement of / 
at /'. The displacement of d may then be found by draw- 
ing lines parallel to 4 and 2 1 through e' and /' and erecting 



Art. 3.] DEFORMATIONS OF A BRIDGE-TRUSS. 

Table I.* 



^T^ 





Total 

Stress in' 

Member 

in Pounds. 


Intensity 




Jl 




Total 


Intensity 




Jl 


Ih* 


01 Stress 


Length 


Defor- 


0) 


Stress in 


ol Stress 


Length 


Defor- 


1 


in Lbs. 
per Sq. 


in 
Inches. 


mation 
in 


J3 

s 


^ Member 
in Pounds. 


in Lbs. 
per Sq. 


in 
Inches. 


mation 
in 


s 


Inch. 




I.iches. 


g 




Inch. 




Inches. 


I 


+ 373.300 


+ 12,000 


360 


+ -154 


8 

7 
6 


+ 3 73,3^0 


+ 9600 


360 


+ -123 


2 

3 


+ 480,000 


" 


" 


" 


r 4~c, 0: 


" 


' 


" 


4 


+ 540,000 


' ' 


" 


' ' 


5 


■f 540, ODC 


' ' 


' ' 


' ' 


9 


— 502,300 


— 9, 000 


472 


-•152 


16 


~S°-,io 


— 7200 


472 


— . 122 


10 


— 501,000 


- 9,500 


376 


-.128 


15 


— 501,00: 


— 7600 


376 


— . 102 


II 


-544,800 


— 10,000 


363 


— . 129 


14 


-544,^0 


— 8000 


363 


- -093 


12 


— 576,000 


* ' 


360 


-.129 


13 


-576,0^0 


' ' 


360 


" 


17 


+ 84,000 


+ 9,000 


324 


+ .104 


29 


+ 84,00c 


+ 7200 


324 


+ .083 


18 


+ 143-500 


+ 10,000 


472 


+ .169 


28 


+ I4i.5cr 


+ S000 


472 


+ -I3S 


19 


— 12,000 


— 1,000 


432 


- .015 


27 


— I2,C0C 


— Soo 


432 


— .012 


20 


+ 93.720 


+ 7.400 


562 


+ .148 


26 


+ 93.72c 


+ 5920 


562 


+ .118 


21 


+ 12,000 


+ 1,000 


480 


+ .017 


25 


+ i?,occ 


+ 800 


480 


+ .014 


22 


+ 60,000 


+ 4,800 


600 


+ . 102 


24 


+ 60,000 


+ -,840 


600 


+ .082 


23 


— 12,000 


— 1, 000 


480 


- .017 




— I-', 000 









£=28,000,000 pounds per square inch. 

* It will be noted that, the right half of this table is similar to the left half, 
but that the intensities of stress for that part of the table are but 80 per cent, of 
those shown for the members indicated in the left half. This was done in order 
to avoid symmetry of the displacement diagram and would not, in general occur 
in practice. 

perpendiculars at their ends. The displacement of k is 
next found by drawing through /' and d' lines parallel to 
II and 20 respectively, equal in magnitude and in the 
proper direction to represent their deformations and 
erecting perpendiculars at their ends. Panel points c, j, b, 
and a are then considered in that order. 

a' shows the final displacement position of a, the left- 
hand end of the truss. The displacements of the panel 
points of the right half of the truss are not shown in detail, 
in order to avoid confusion of the diagram, but they would 
be found in a precisely similar manner. The diagram 
shows only the displacement of the right end i at i'. 

The figure has so far been drawn upon the assumption 
that the point m is fixed in position and the direction of 



172 DEFORMATION OF TRUSSES. [Ch. V. 

me fixed. Actually, however, it is the point a which 
remains fixed in position and the point e is displaced to 
the amount a'e' . The diagram also shows the displace- 
ment of i to be i'e', whereas the point can move only in a 
horizontal line. The distorted figure, now considered rigid, 
must therefore be so rotated about the point a that the 
point i falls in the line in which the end rollers move. 
By the methods of the previous article, a' at once becomes 
a", and it suffices merely to find the displacement of one 
other point in order to locate all the other displacements. 
The point i moves in a line at right angles to ai when the 
truss rotates about a. In Fig. lo, therefore, the line a"i" 
is drawn at right angles to ai, and its intersection with 
i'i", drawn horizontal, determines the final displacement 
of i. i'i" is drawn horizontal, for that is the only possible 
direction of its displacement. The points a" and i" now 
furnish two points for the diagram from which the displace- 
ment of all other points in the figure, as it is rotated about 
a, may be found. It is simply necessary to draw upon 
a"i" as a lower chord a reproduction of the main truss, 
a"j"k"l" . . . i" , every line of which must be at right angles 
to the corresponding line of Fig. 9. 

The final displacement of every point is now found as 
the resultant of the two previous displacements, these latter 
always being measured in the direction of the second 
primed letter toward the first primed letter. For instance, 
the displacement of the point e is downward from e" to e'. 

In general it is sufiicient to determine merely the 
vertical projections of the displacements or the deflections 
of the panel points carrying the moving load, for these 
deflections indicate the extent to which the truss must 
be cambered. They may be found by projecting the posi- 
tions of the primed letters e' , d', c', etc., to the left until 
they intersect at ei, di, ci, etc., verticals dropped from the 
panel points. The open polygon obtained by connecting 



Art. 4] DISPLACEMENT DIAGRAM FOR A THREE-HINGED ARCH, i 73 

these points may be closed by a line aii2 in the manner of 
closing a funicular polygon. The deflection of any panel 
point is then represented by the intercept in the polygon 
directly below that point, e^ei, for instance, being the deflec- 
tion of the point e. 

Art. 4. — Displacement Diagram for a Three-hinged Arch. 

Fig. 12 represents a three-hinged arch for which it is 
desired to obtain the displacements of all the panel points 
when the structure carries load. The stresses, the sectional 
areas, the lengths, and the deformations of the members 
are not given in tabular form, but the strain to which each 
member is subjected is shown to scale in the truss diagram 
by a heavy line on each member, the minus sign near 
each member representing a compressive strain, while the 
plus sign indicates an elongation. 

Treating first the left half of the arch, it is assumed that 
kd suffers no change in direction and that the point d 
remains fixed in position. Under that assumption the 
usual displacement diagram, Fig. 13, for the left half of 
the arch may be drawn. A line parallel to i is drawn verti- 
cally downward from d' (the pole) to represent the com- 
pressive strain in kd. At d' a line parallel to 2 is then 
drawn to indicate the displacement of ;'. The displacement 
of c is then found by drawing lines through ;' and d' parallel 
to 4 and 3 respectively and finding the intersection of the 
perpendiculars erected at the extremities of these lines. 
The displacements of the panel points i, b, h, and a are 
then found by treating the strains in the order in which 
the members are numbered. 

The right half of the arch is then to be treated in precisely 
the same manner, but it is sufficient to indicate in Fig. 1 3 
simply the displacement of the point g, namely g'. It is 
now necessary to determine the rotation of the two halves 



^74 



DEFORMATION OF TRUSSES. 



[Ch. V. 



h - 9 i + r, ./ I 



I M + ':> 20 + ^i 




Fig. 12. 




Fig. 13. 



Ai;t. 4 ] DISPLACEMENT DIAGRAM FOR A THREE HINGED ARCH, i 7 5 

c f the arch about the points a and g as centres respectively 
in such manner that the point d may coincide for both 
halves. 

Since the two parts of the structure rotate respect- 
ively about the fixed points a and g, the points a' and g' 
of Fig. 13 become at once a" and g" . Treating the left 
half of the arch first, it is seen that d rotates at right angles 
to the line ad, connecting a with d. In the displacement . 
diagram (Fig. 13) the path of this rotation is indicated by 
a"d", a line drawn through a" at right angles to ad. 
Similarly the right half of the arch must rotate about g, 
and the path: of the rotation of d about g is represented by 
a line drawn through g" at right angles to gd of Fig. 12. 
The intersection of these lines fixes at once the final dis- 
placement of ! <i at d" . Since two final points of displace- 
ments have been found for each half of the arch, the dis- 
placement of all other points may be. found by drawing 
a figure between d" and a" precisely similar to the right- 
hand half of the arch, but with every line at right angles to 
it. It is indicated in Fig. 13 by d"k"h' a" . The displace- 
ment of any point in that part of the structure is obtained 
by a line connecting the double-primed with the single- 
primed letters, the displacement of h for instance being 
h"h'. 

In order to avoid confusion the displacement of the 
points on the right half of the figure are not shown. There 
would be required simply a figure drawn between g" and 
d" similar to the right half of the arch, but with every 
line at right angles to it. 

Displacements due to changes of temperature are 
treated in precisely the same manner as the displacements 
due to stress deformations. The changes in length caused 
by the stresses are merely replaced by the changes in 
length caused by the change in temperature. 



i7Sa 



DEFORMATION OF TRUSSES. 



[Chap. V- 



Art. 5- — Displacement Diagram for Truss with Sub-divided 

Panels. 

In Fig. 1 4 is shown a Williot or displacement diagram, 
for dead load only, for a truss with sub-divided panels, 
having a span length of 350 feet; the complete design of 
this truss is given in Chapter VI. 

Table I. 





Total Fixed 


Area of Cross- 


Length 


Distortion of 


Member. 


Load Stress 


section. 


in 


Member 




in Pounds. 


Square Inches. 


Inches. 


la Inches. 


h.L, 


+ 234,000 


46 


35° 


+ 0.0635 


L,U 


+ 234,coo 


51 


350 


+ .0572 


L,L, 


+ 340,000 


71 


3.=iO 


+ -0525 


L,L 


+ 382,000 


81 


3 SO 


+ -059 


LU 


+ 382,000 


83 


35° 


+ -0575 


L,L 


+ 424,000 


90 


35° 


+ -0589 


L,U, 


-371,000 


Q7-23 


556 


- .0758 


u,u. 


-356,000 


83 25 


366 


- .056 


U.Us 


— 400,000 


90.25 


366 


- .058 


u^u. 


— 45 3 000 


99.1 


354 


- .0578 


UJJr 


-453,000 


99.1 


354 


- .0578 


u,u, 


— 448 000 


99.1 


350 


— .o:;6 


U,Li 


+ 3^.000 


17.9 


432 


+ .02S4 


U,L, 


+ 168 000 


34-0 


556 


+ .0982 


UL, 


— 98,000 


39-4 


540 


— .04.63 


TJL 


+ 78,000 


20. 1 


643 


+ -0893 


u,u 


— 32,000 


19.4 


648 


- .0382 


U,M, 


+ 90,000 


25-4 


477 


+ .0603 


W^ 


— 20,000 


12.9 


363 


— .0201 


M^Li 


+ 33,000 


24-5 


324 


+ .0156 


¥A? 


+ 5 7,000 


20. 1 


477 


+ .0483 


MU, 


+ 30,000 


12 


556 


+ .0496 


Vr'h 


— 7,000 


19.4 


756 


— .00976 


U,M, 


+ 36,000 


193 


515 


+ -0343 


V,M. 


— 20,000 


12,9 


378 


— .0209 


M,L, 


+ 33.000 


14.4 


378 


+ .031 



Coefficient of EIasticity= 28,000,000 ibs. per sq. inch. 

In Table I are shown for each member the total dead 
load stress, the area of cross-section in square inches, and 



Art. 5,] DISPLACEMENT DIAGRAM FOR TRUSS. 17 sd 

the length in inches. In the last column there have been 
calculated the changes in length of the various members due 
to dead load, the coefficient of elasticity being assumed at 
28,000,000 lbs. per square inch. These changes in length 
are written along the members of the truss in Fig. 16. 

The Williott diagram is begun by assuming panel point 
o as fixed in position at 0' and the displacement of p with 
regard to is taken in the direction of op, so that ths new 
position of p becomes p'. The displacements of all other 
panel points is then found with reference to that displace- 
ment. Since the member op suffers an elongation of .031", 
the line o'p' must be drawn downwards to represent that 
displacement. 

The displacement of the panel point / at /' may next be 
found by drawing a line from p' parallel to pi with a length 
of .0589 inches. 

The displacement of m at nt' is next found by drawing 
a line Vm' .00976" long parallel to the lower chord member 
Im and erecting a perpendicular at the extremity of that 
line. There is then drawn at 0' a line parallel to om, having 
a length of .0343", and a perpendicular erected at its ex- 
tremity. The intersection of the two perpendiculars thus 
drawn, determines the position of the point m'. 

In precisely the same manner the displacement of other 
panel points may be found; j might be determined next, 
and then h, k, g, etc. ; after the left half of the truss is com- 
pleted, the same methods of construction are employed for 
the right half of the truss. 

The final points to be located will be the two ends of the 
truss, a and ai at a' and ai respectively, a/ being found 
0.68 inches to the right and 0.99 inches above a'. 

Since the point a is fixed in position, its true displacement 
must be zero and the point marked a' must be a" or the 
origin about which to rotate the figure of the truss in order to 



T/Sc DEFORMATION OF TRUSSES. [Ch. V. 

determine all other displacements ; since the point ai rests 
on rollers which can move only in a horizontal line, the true 
displacement of ai is found by drawing a horizontal line 
through a/ until it intersects a vertical erected at a"; that 
intersection is marked a/', and the distance from a/' to ai' 
measures the true and final displacement of the point a^ ; 
viz., 0.68 inches to the right 

As in all previous cases the line a"a\' is the base line 
upon which there is now to be drawn a reproduction of the 
figure of the truss, the line a" ax" being used as the lower 
chord line ; all panel points of this truss are marked with 
the second primed letters. 

The final and true displacement of any point of the 
truss under the dead load may now be found, since it is that 
displacement which is measured from the point marked by 
the second primed letter to the point marked by the single 
primed letter ; for instance, the point p is actually displaced 
by the distance p" p' , or a vertical displacement downwards 
of 1.57 inches and a horizontal displacement toward the 
right of 0.34 inches. 

As in the previous examples of the Williott diagram, the 
strains or distortions of the members of the truss are shown 
in the diagram as heavily shaded lines, and they are marked 
with the letters indicating to which member of the truss 
they pertain ; the perpendiculars erected at the extremities 
of the heavier lines are shown with a light line. 



CHAPTER VI. 

THE DETAILED DESIGN OF A RAILROAD BRIDGE. 

The design of the bridge will be based on the following 
data ; * in engineering practice these would be furnished by 
the railroad company for which the bridge is to be built: 

Length of span between centres of end pins. Panel length, 

29' 2", and number of panels, 12 350' o" 

Distance between centres of trusses 18' o" 

Distance between centres of stringers 7' 6" 

The centre height I/5L5 is 63' o" 

The intermediate height UJL^ is 54' o" 

The end height t/jL, 36' o" 

The bridge is a through one, for single track, and 
the form of truss is of the type known as the broken 
upper chord with subdivided panels, as is clearly shown 
on the stress sheet, Plate I. As specified by the railroad 
company, all of the material of the bridge is to be of medium 
steel; all the rivets are to be | inch, except that |-inch 
rivets may be used in the flanges of 12 -inch channel posts, 
in horizontal struts at the middle of posts, and in the over- 
head lateral or sway bracing. The structure is to be 
designed according to Cooper's specifications of 1896. 

* The authors are indebted to the American Bridge Company of New York 
for the stress sheet and detail drawings of this bridge; the original design was in 
charge of Mr. O. E. Hovey, Engineer of Design. 

176 



Art. I.] STRESSES IN THE STRUCTURE. 177 



Art. I. — Stresses in the Structure. 

The stresses in the structure are caused by its own 
or dead weight and by the live or moving load. The dead 
load may be divided into two parts, the weight of the floor 
system consisting of rails, ties, track-fastenings, stringers, 
and floor-beams, and the trusses and bracing. . Excerpts 
from Cooper's specifications relating to these loadings are 
the following: 

§ 2j. All the structures shall be proportioned to carry the 
following loads: 

1. The weight of metal in the structure. 

2. A floor weighing 400 pounds per linear foot of track 
to consist of rails, ties, and guard timbers only. 

These two items, taken together, shall constitute the dead 
load. 

J. A ''live load " on each track, supposed to be moving in 
either direction, consisting of 2 ''consolidation''' engines, 
coupled and followed by a train load, distributed as shown on 
diagram E 40* (or 100,000 pounds equally distributed on two 
pairs of driving-wheels, spaced seven and a half feet, centre to 
centre). -\ 

The maximum stresses due to all positions of either of the 
above " live loads'' of the required class, and of the " dead 
load," shall be taken to proportion all parts of the structure. 

§ 27. Variation in temperature, to the extent of ijo° F., 
shall be provided for. 

§ 28. When the structures are on curves, the additional 
effect due to the centrifugal force of trains shall be considered 
as live load. It will be assumed to act 5 feet above base of rail, 

* For a diagram of locomotive E 40, see page 100. 

t This part of the specification is usually employed in determining the maxi- 
mum bending moments and shears in the floor system. In the example under 
consideration this alternative loading has been neglected. 



178 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 

aitd ivill be computed for a speed of jo - 2d miles per hour; 
d being the degree of curve. 

\ 2g. All parts shall be so designed that the stresses coming 
upon them can be accurately calculated. 

§ 28 does not apply to the structure under .consideration, 
and the requirements of § 27 may be fulfilled by placing 
one end of the bridge upon a nest of movable rollers. 



Art. 2. — Determination of Dead-load Stresses. 

The weight per linear foot of the trusses, not including 
the floor system but including all the lateral and cross 
bracing, may be approximately estimated by means of the 
following" formula; W = al, where W is the weight per 
linear foot of both trusses and bracing in pounds, a is a 
numerical coefficient, and / is the length of span in feet. 
In the bridge under consideration a may be taken as 
7.75. The weight of the two trusses then becomes 2700 
pounds per linear foot, or 39,380 pounds per panel per truss. 
This weight must be equally divided between the upper 
and lower panel points. The specifications require the 
weight of track to be taken at 400 pounds per linear foot, 
while the remainder of the floor system may be assumed 
to weigh 500 pounds per linear foot; hence these loads 
become 5830 and 7290 pounds per panel per truss respect- 
ively. They act at the lower panel points. 

The upper panel-point loading is then 19,690 pounds, 
but the lower panel-point loading becomes 32,810 pounds. 

The dead-load stresses are easily determined by graphic 
methods as shown in Fig. i ; for that figure only those 
web members shown in full lines are assumed to act, the 
dotted members acting as counters.* It is unnecessary to 
explain the details of the figure ; but to insure the accuracy 

* See Chap. II, Art. i8, for the treatment of subdivided panels. 



Art. 2.] DETERMINATION OF DEAD-LOAD STRESSES. 



179 



of the graphic diagram it is important to check analytically 
the last stress found. The stress in Ld-&, for instance, is 
so found by passing a section through U^Uq, UJVL&, and 
LfiLe and taking the centre of moments at U s- 













Y 


y' 


N tj V "^5 0-448 C^6 


J PaD:::*^— T- / 


X-* z 


z' / 


>^ G , ^2 p=:i^ 


fXx Oo 


1-/^ 


\^. f 




) u^^^ 


V 




y/ 


\ c> 


/ 


/ 

A / 

,V B " 

/D+IH ^ 


2) + 234 \ 


\ " 
\ 1 

H \ 

D + 340 \ 


L 

y + 

,-^0+382 "^ 


(Mi T 

B + 382 \^ 


Xl 

/b+424 


\ u' 



Assumed Dead Load: 

Weigkt of Track = 400*p. lin. ft,= 5830 "^p. panel p. truss. 

" " Floor =500*" " 

" " Trusses=270O*" " 
TTpper Chord panel load=l%90* 
Lower " " " =19690+13120=32810* 
D indicates dead load stress in thousands of pounds. 




Fig. I. 



The compression stresses in L3M4: and M4U5 are found 
by assuming U^M^ not to exist and are quickly determined 
by the method of sections. 



l8o DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 



Art. 3. — Determination of Live-load Stresses in the Chords. 

The position of the moving load causing the greatest 
bending moment at any point in a truss or beam is found 
by the following criterion,* it being supposed for con- 
venience that the train passes along the truss from right 
to left with Wi as the front load: 






(I) 



In eq. (i), /' represents the distance from the left end 
of the truss to the centre of moments, i.e., to the point 
in question, and / the total length of the truss; 
W^i + M/2 • . • Wn the sum of the loads on /' and 
Wi + W2 . . . Wn the sum of the loads on I. 

By the method of sections the stress in a chord member 
is found by dividing the maximum bending moment at 
the proper panel point by the normal distance (or lever- 
arm) from the panel point to the chord member in question. 
This method is directly applicable to the truss considered, 
although the modification furnished in Chap. II, Art. 20 is 
necessary in the case of the chord members in the sub- 
divided panels. 

] By drawing a reaction influence line both the truss 
reaction and the bending moment at any point for any 
position of the moving load may be at once measured; 
the diagram should be made on a large scale in order to 
attain the highest practicable degree of accuracy. 



* See eq. (6), Chap. II, Art. 6. 



Art. 4.] LIVE-LOAD STRESSES IN IVEB MEMBERS. i«i 

TABLE I. 



Member. 


Wheel-load 
Number. 


At Panel Point. 


Stress in Pounds. 


Remarks. 


LoL, 


5 


L, 


+ 333.000 


Train advancing 
from right to left 


L^L^ 


S 


L, 


+ 333,000 


do. 


L,Ls 


13 


L, 


+ 480,000 


do. 


L,L, 


17 


L, 


+ 537.000 


do. 


L,L, 


17 


U 


+ S37.000 


do. 


L,L, 


23 


L, 


+ 590,000 


do. 


L,U, 


5 


h 


— 528,000 


do. 


U,L\ 


13 


L, 


— 502,000 


do. 


U^U, 


17 


L, 


— 562,000 


do. 


U,Ui 


20 


L, 


— 631,000 


"1 Criterion of the 


u,u. 


20 


L, 


— 631,000 


subdivided 


u,u. 


26 


4 


— 624,000 


J panel 



By applying the criterion, eq. (i), and completing 
these diagrams, the results shown in the preceding table 
will be obtained. The table exhibits both the positions 
of the concentrations for the greatest bending moments 
and the greatest stresses in the various chord members. 

The uniform load is treated as a uniform series of 
loads of 25,000 pounds each; i.e., load No. 23 in the table 
indicates the fifth load of this series. 



Art. 4. — Determination of Live-load Stresses in Web Members. 

The methods of Chap. II, Art. 16 were used to determine 
the position of the loading causing the maximum stresses 
in the web members of this truss, the modifications neces- 
sary for subdivided panels being given in Art. 19 of the 
same chapter. 

The application of those methods gives all the* results, 
both as to positions of loading and the greatest web stresses, 
exhibited in Table II: 



1 82 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 

TABLE II. 



Member. 


Wheel-load Number. 


At Panel Point. 


Stress in 
Pounds. 


Remarks. 


U,L, 


4 


L, 


+ 257,000 


Train advancing 
from right to left 


U^L, 


3 


L, 


— 161,000 


do. 


U^L, 


3 


i, 


+ 160,000 


do. 


U,L, 


3 


L, 


— 103,000 


do. 


U,M, 


3 


Lf, for panel L^L^ 


+ 204,000 


do. 


M,L, 


5 


Li,, for panel Z3L5 


+ 168,000 


do. 


U,M, 


When LfMf is max. 




+ 62,000 


do. 




3 


L3, for panel L^L^ 


+ 104,000 


Train advancing 
from left to right 


M,L, 


S 


L3, for panel I.3L5 


+ 82,000 


do. 


U,L, 


3 


L, 


— 79,000 


Train advancing 
from right to left 


U,M, 


3 


Le, for panel L^L^ 


+ 173,000 


do. 


M,L, 


3 


L5, for panel L5L5 


+ 136,000 


do. 



The members UiLi, M^L^, and MqLq are simply hangers. 
They carry in tension the maximum floor-beam load, which 
occixTs when wheel No. 5 is directly at the hanger. 

The members UiMi and UqMq are simply supporting 
struts; they carry in compression the dead load assumed 
to exist at their upper ends, but they carry no live load. 

The horizontal struts MzM^ and M^M^ are intended 
simply to prevent flexure of the long column into which 
they frame. They are subject to no direct stresses of 
either tension or compression. 

The determination of the maximum tension in the ver- 
tical members UzL and U^L^ has already been treated in 
full in Art. 21, Chap. II. 



Art. 5.] DETERMINATION OF IVIND STRESSES. 183 



Art. 5. — ^Determination of Wind Stresses. 

Cooper's specifications prescribe: 

" § 24. To provide for wind stresses and vibrations . . . 
the bottom lateral bracing in through bridges, for all spans 
up to joo feet, shall be proportioned to resist a lateral force of 
4^0 pounds for each foot of the span; joo pounds of this to 
be treated as a moving load, and as acting on a train of cars, 
at a line 8.j feet above base of rail. 

" . . . the top lateral bracing in through bridges for all 
spans up to joo feet shall be proportioned to resist a lateral 
force of ijo pounds for each foot of span. 

''For spans exceeding joo feet add in each of the above 
cases 10 pounds for each additional jo feet of span." 

In accordance with these specifications, the asstimed 
wind load for the bottom chord will be 470 poiinds per 
linear foot, and for the upper chord 170 poirnds per linear 
foot. 

Upper Lateral Bracing. 

The stresses for the upper chord lateral bracing are 
found in the most expeditious way by analytical methods.* 

The bracing may be considered a truss of 12 panels 
lying in a horizontal plane with its abutments at the feet 
of the end posts. The loading is fixed in position and 
equal to 29.1X170=4950 pounds per panel, the reaction 
at the end post being 29,700 pounds. The secant of the 
angle of inclination between the. web members and the 
perpendicular to the plane of the. main truss is 1.9. The 
stresses in the web members are then found by mtiltiplying 



* For the treatment of wind bracing, as exemplified in this chapter, the reader 
is referred to the authors' "Metallic Bridges." 



i84 DETAILED DESIGN OF A RAILROAD BRIDGE^ [Ch (l. 

the shear in the respective panels by this secant, ^nd 
they are given on the stress sheet. It should be noted 
that in this treatment the panel of the portal is considered 
a part of the wind bracing, although the true lateral brac- 
ing in that panel is replaced by a portal frame. This 
involves no error in determining the stresses of the webbing 
of the system, and is on the side of safety for the chord 
members. The stresses in the chord members are easily 
found by the method of sections. 

Lower Lateral Bracing. 

The lower lateral bracing is designed for a moving load 
of 300 pounds per lineal foot and a fixed load of 170 pounds 
per lineal foot. The stresses in the chord members of that 
system are easily found by multiplying the stresses in 
the chord members of the upper lateral system by the ratio 

between the loads of the two systems, i.e., = 2.76. 

The sum of the fixed and moving loads is taken, since the 
truss must be entirely covered by the moving load for max- 
imum chord stresses. The resulting stresses are shown on 
the stress sheet. 

The stresses in the webbing are found by taking the 
stationary load the same as for the upper chord bracing 
and combining its effects with those due to the moving 
load. The following examples illustrate the method 
employed : 

Weh Member in Panel LqL\. 

The stress in this member has its maximum value when 
the entire truss is covered by the moving load. The reaction 
then becomes 52,500 pounds, and the shear in the panel 
is 52,500 — (29.1 X3oo)/2 =48,150 pounds. The stress is 
then 48,150X1.9=91,500 pounds. To this must be added 



Art. 6.] DESIGN OF LOlVER CHORD MEMBERS. 185 

the stress due to the stationary wind load, viz., 5100 
pounds. The sum of these two quantities may then be 
assumed to be carried equally between the two intersecting 
web members within the panel, causing compression in 
the one and tension in the other member. 

The resultant stress for each is therefore ^(91,500 + 
51,000) =71,250 pounds. 

Web Member in Panel L1L2. 

For this member, ^^/n of the bridge, or 318. i feet, should 
be covered by the uniform moving load.* The shear in the 
panel is then 43,400 — 3620=39,780 pounds, and the 
stress is 75,500 pounds. The stress due to the stationary 
load is 42,000 pounds; hence the final stress in the member 
is i(75, 500 + 42,000) =58,750 pounds. 

The other web members of this lateral system are 
treated in precisely the same manner. 



Art. 6. — Design of Lower Chord Members. 

The specifications which apply to the design of the 
lower chords are the following: 

§ JO. All parts of the structures shall be proportioned in 
tension by the following allowed unit stresses: 

Bottom chords, main diagonals, counters, and long verticals 
(forged eye-bars) — for live loads 10,000 pounds per square 
inch and for dead loads 20,000 pounds per square inch. 

§ 3<5. The stresses in the truss members or trestle posts 
from the assumed wind forces need not be considered except 
as follows: 

1st. When the wind stresses on any member exceed one 
quarter of the maximum stresses due to the dead and live loads 

* The student may check this by the method of influence lines. 



i86 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VL 

Upon the same member. The section shall then be increased 
until the total stress per square inch will not exceed by more 
than one quarter the maximum -fixed for dead and live loads 
only. 

Member LsLq. 

The stresses in this member, as taken from the stress 
sheet, are: 

Dead-load stress = +424,000 pounds. 
Live-load stress =-1-590,000 " 
Wind stress =±402,000 " 

Neglecting the wind stress, the area of cross-section 
required would be 80.2 square inches, as shown: 

424,000 H- 20,000 = 21.2 square inches. 
590,000 H- 10,000 = 59.0 

80.2 

The actual intensity of stress in the member would 
then be (424,000 + 590,000) h- 80.2 ==12,600 pounds per 
square inch. The unit stress due to wind would then be 

402,000 . 

—X =5010 pounds, 

which exceeds by 5010 — 3150 = 1860 pounds, one quarter 
of the unit stress due to the dead and live load. The 
sectional area of the chord member, therefore, must be 
increased until the intensity of stress does not exceed 
12,600 X|- = 15,750 pounds. The combined dead, live, and 
wind loads is 1,416,000 pounds. Hence the required area 
of cross-section will be 

1,416,000 „ . , 

=oQ-0 square inches. 

15.750 ^^ ^ 



Art. 6.] DESIGN OF LOIVER CHORD MEMBERS. 187 

The member will then be composed of six eye-bars, each 
8 inches deep by i| inches thick, the area of whose com- 
bined cross-section will be 90 square inches. 

Other lower chord members are treated in exactly the 
same way, and it will suffice to consider one more only. 

Member L^L^. 

Dead-load stress= -1-382,000 lbs. @ 20,000 lbs.= ig.i sq. in. 
Live-load stress = 4-537,000 lbs. @ 10,000 lbs. = 53.7 " 

72.8 " 

Wind stress = -I- 356,000 lbs. 



Total stress = -1-1,275,000 " 

Neglecting wind stress, the average stress in the member 
would be 

382,000 -f 537,000 ^ . . , 
^ = 12,620 pounds per square mch. 

Adding 25 per cent, to this stress on account of wind, the 
final section required becomes 

1,275,000 

—. 2 \ \ =80.8 square mches. 

(12,620 + 3,154) 

The member will be composed of six eye-bars, each 
8 X itt inches, whose combined area will be 81 square inches. 

The lower chord members in the first two panels of the 
truss are subjected to a more sudden loading than those 
in the other panels ; it is therefore customary to form these 
members either of shapes suitable to resist compression, 
such as channels, or stiffen them by latticing the eye-bars 
together. In the present case the latter plan will be 
adopted; the two centre eye-bars, four being necessary, 
will be tied together by 2^Xi-inch lattice bars. 



1 88 DET/1ILED DESIGN OF A RAILROAD BRIDGE. [Ch.VL 



Art. 7. — Design of Upper Chord Members. 

The following specifications apply to the design of the 
tipper chord : 

§ J J. Compression members shall be proportioned by the 
following allowed unit stresses: 

Chord segments: 

I 
P = 10,000 — 4j— for live-load stresses; 

P = 20,000 — go— for dead-load stresses. 

End posts are not to be considered chord segments. All 
posts of through bridges: 

P= 8500 — 45— for live-load stresses; 
P = i'/,ooo — go— for dead-load stresses, 



in which expressions P is the allowed stress in pounds per 
square inch of cross-section, I is the length of compression 
m,ember in inches; r is the least radius of gyration of the 
section in inches. 

No compression member, however, shall have a length 
exceeding 12 j times its least, radius of gyration. 

§ yo. The unsupported width of plates subjected to com- 
pression shall not exceed jo times their thickness, except cover- 
plates of top chords and end posts, which will be limited to 
40 times their thickness. 

§ go. In compression chord sections the material must 
mostly be concentrated at the sides, in the angles and vertical 
ribs. Not more than one plate, and this not exceeding \ inch 



Art. 7.] DESIGN OF UPPER CHORD MEMBERS. 189 

in thickness, shall be used as a cover-plate, except when neces- 
sary to resist bending stresses or to comply with § 70. 



Member U^Uq. 

Dead-load stress = — 448,cxdo pounds. 
Live-load stress = —624,000 " 

It will now simplify the design work to reduce the dead- 
load stress to the equivalent live-load stress; for this 
member the live-load stress, plus one half the dead-load 
stress, becomes 848,000 pounds. 

The allowed unit stress is not a constant quantity for 
compression members, but it depends upon the radius of 
gyration of the cross-section of the member. It becomes 
necessary, therefore, in designing, to assume some trial 
value for the radius of gyration. In the rectangular or 
box form of section adopted for upper chord members, 
it will usually be found that the radius of gyration has 
a value of about 0.4 the depth of the side plates, the depth 
of the member being taken slightly greater than would be 
the case in a vertical column on account of the bending 
induced by its own weight. This depth is usually about 
one-twelfth to one-fifteenth the length. 

In the case of U^Uq, the depth of the member may be 
taken as 28 inches; the assumed radius of gyration be- 
comes 1 1.2, and since / is 350 inches, the assumed unit 
stress becomes 



P = 10,000 — 45 =8594 pounds per square inch. 



The approximate area required is therefore 
848,000-^8594 = 98.7 square inches. 



190 



DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 



A trial cross-section must now be formed from angles 
and plates of about this area, and tested to ascertain 
whether the assumed radius of gyration must be modified. 
The upper angles in the trial section should be made as 
light as possible, so that the cover-plate may be largely 
balanced by the heavy bottom angles. Flat bars are 
frequently added to the horizontal legs of the lower angles 
for the same purpose. In this manner the centre of gravity 
of the section may be brought down sufficiently near to 
mid depth to give the space needed inside the chord for 
the eye-bar heads, the pin axis being made to pass through 
the centre of gravity of the section. At the same time 
there is gained the incidental but important advantage of 
an increased moment of inertia. If the chord were subject 
to no bending from its own weight, the axis of every pin 
should pass through the centre of gravity' of the section. 
It may be shown * that this flexure cannot be satisfactorily 
met and neutralized by the direct stress, particularly if 




chord-section u5u5 
Fig: 2. 



the chord is considered as continuous. It is best there- 
fore to reduce the bending stresses by making the chord 
depth as great as possible and placing the axis of the pin 
either through the centre of gravity of the section or but 
slightly below it. 

* See Art. 6g, Burr's "Course on Stresses in Bridge and Roof Trusses." 



Art. 7.] DESIGN OF UPPER CHORD MEMBERS. lyi 

Let the following section (shown in Fig. 2) be tested: 



Sq. In. 



1 cover-plate 30 X |" =18 

2 angles 4"X3i"xA"= 7 

2 web plates 28"X Y' = 28 

2 " " 28"XA" =24 

2 angles 5"X4"X^" = 9 

2 bars 6"Xi" =10 



99-05 

The position of the centre of gravity of this section must 
first be determined by taking moments about the upper 
edge. The following calculation will determine its position: 

Moment of cover-plate 18.75X 0.32= 5.9 

" " upper angles 7.80X 1.91= 14.9 

" " web plates 52.50X 14.63 = 768.0 

" " lower angles 9.50X27.51 = 261.8 

" " bars 10.50X29.07 = 306.0 



Total= 1,356.6 

Dividing this sum by the total area of the section deter- 
mines the distance of the centre of gravity from the upper 
edge, i.e., 

1356.6 



99-05 



= 13.8 inches. 



The moment of inertia of the section must now be found 
about this centre of gravity as an axis: 

Cover-plate, ^^^^y~ + [i&-7 5^(^3-5)']= 3>42o.6 
Upper angles, 2Xs.7i-f-[7.8oX(ii.9)^]= 1,114.4 

Web plates, ^-^^^^j^:^ + [52. 50X {.&)']= 3,463.6 

Lower angles, 2X6.27 -|-[9.5oX (13.7)^= 1, 794-5 

Bars. "^^f^^^' +[io.5oX(i5.3)']= 2,460.7 

Total= 12,253.8 



192 DET/IILED DESIGN OF A RAILROAD BRIDGE. [Ch. VL 

The radius of gyration is equal to the square root of the 
quotient obtained by dividing the moment of inertia by 
the area, that is, 



where r = the radius of gyration of section, 

/ = moment of inertia of section, 
and A =area of section. 
In the present case 



12,253.8 . , 

r = \\ = 11.1 mches. 

99-05 

This is so nearly the value assumed that it will not be 
necessary to repeat the calculations with the true value of r. 
If there had been a considerable difference between the 
assumed and the actual values of r it would have been 
necessary to repeat the calculations with such changes 
in the trial section as would furnish satisfactorily correct 
results. 

It must be determined whether the moment of inertia 
may not be smaller about the vertical axis of symmetry 
of the section; if this be the case, this smaller value of r 
must be used to determine the allowed intensity of stress. 
The calculations for this moment of inertia are as follows: 

Cover-plate, — ^ ^"^ -= 1,405 

12 

Upper angles, 2[4.o7 + (i2. 53)2X3. 9]= 1,236 

rCJL5.")3y28 ~\ 

Webplates, 2 ^"^^ — + 26.25X(ii.o3)M = 6,404 
Lower angles, 2[ii. 26+ (13.12)^X4.75]= 1,656 

Bars, ^[ '^^^^/'^^' + 5-^5X(i3-5)^]- i,944 

Total =12,645 



Art 7.] DESIGN OF UPPER CHORD MEMBERS. 193 

This moment of inertia is larger than the one found 
for the horizontal axis, and it need not therefore be further 
considered. 

The distance from back to back of the lower and upper 
angles will be made 28^ inches, being slightly greater than 
the depth of side plates in order to prevent their inter- 
ference with the cover-plate or lower bars during manu- 
facture. The axis of the pins will be placed 14J inches 
from the upper edge of the cover-plate, thus allowing 
a small counter moment from the direct stress due to a 
lever-arm of 0.7 inch. The other upper chord members 
are treated in precisely the same manner. The stress 
sheet, Plate i, shows in sufficient detail the forms of cross- 
section adopted. 

End Post LqUi. 

The end post is designed under a different provision of 
the specifications, the allowed intensity of stress for live 
load being found by the following formula: 

^ = 8500-45—. 

The length of the end post between pins is 558 inches, 
but it may be assumed that the effective length is only 
one half of this, since the post is supported at its centre 
in one direction by an inclined strut, and in the direction 
at right angles by the portal bracing. Assuming ^ = 279 
inches and r = ii.i inches, the allowed intensity of stress 
becomes 7370 pounds per square inch. 

Dead-load stress = —371,000 pounds 

Live-load stress = — 528,000 " 

{\ dead -t- live) load stress= — 713,500 " 

Therefore — ^-^^ — =g6.g square inches required. 



194 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 

The stress sheet indicates the section adopted for this 
member. It will be found on examination that the require- 
ments of the specifications concerning the unsupported 
widths of plates in compression have been fulfilled. 

Art. 8. — Design of Floor-beam Hangers. 

UxLu LdU, UMq. 

These members are built from shapes rather than from 
bars with forged ends, in order that the floor-beams may be 
attached to them with the same details that are used at 
other panel points. The following specifications apply: 

§ JO. Floor-heain hangers and other similar members {in 
tension) liable to sudden loading {plates and shapes), net 
section at 6000 pounds per square inch. 

Long verticals in tension {plates or shapes), net section at 
gooo pounds per square inch for live loads and 18,000 pounds 
per square inch for dead loads. 

UiLi, L4M4, and L^Mq are long verticals, and they are 
covered by the second of the preceding provisions. 

Member UiLi. 

Dead-load stress = + 33,000 pounds 

Live-load stress = + 84,000 ' ' 

(I dead -f-live) load stress = -1-100,500 " 
~55§D'-= ii-i? square inches required. 

The following 15 -inch channels spaced 14 inches back 
to back more than satisfy this requirement: 

Two 1 5 -inch channels weighing 32.95 pounds per lin. ft. (^.9.69 
square inches = 19.38 square inches gross or 17.88 
square inches net, held together by single latticing 
2^ X finches. 

This is much more metal than is needed, but the member 
is 36 feet long and requires increased stiffness on that 



Art. 9.] DESIGN OF VERTICAL MAIN IV EB MEMBERS 195 

account. As these same channels must be used for a 
number of the main web members of the truss, they will 
also be used for these long verticals. 

Members L^M^ and LqMq. 

Both of these members are subjected to the same 
stresses, and since they occupy similar positions in the 
truss, they will be designed exactly alike. 

Dead-load stress = + 33,000 pounds 

Live-load stress = -t- 84,000 ' ' 

(h dead -f- live) load stress = -1-100,500 " 
i5g§Q-= 11.16 square inches required. 

A narrower channel than 12 inches cannot be used on 
account of the floor-beam connection ; it will be necessary 
to use two 1 2 -inch channels, each weighing 24.51 pounds 
per linear foot, with a total gross area of 14.42 square 
inches and a net area of 12.86 square inches tied together 
by single latticing 2^Xi| inches. These members will be 
continued vertically upward to form the members M4U4 
and MeUe respectively; their sections are ample to carry 
the compressive stresses imposed. 

Art. 9. — Design of Vertical Main Web Members or Posts. 

U2L2, UsLs, U,L,. 

These main vertical posts are all in compression and 
therefore subject to the following specification: 

§ J J. All posts of through bridges shall be proportioned by 
the following allowed unit stresses: 

P= 8,joo — 4j— for live-load stresses; 
P = 1^,000 — go— for dead-load stresses. 



196 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI.. 

Member U2L2. 

Dead-load stress = — 98,000 pounds 

Live-load stress = — 161,000 " 

(^ dead 4- live) load stress = — 2 10,000 ' ' 

By assuming the width of post at 22 inches, and the 
radius of gyration .35 of this width, the length of the post 
being 540 inches, the allowed intensity of stress becomes 

^ = 8500 — = 5340 pounds per square inch, and the 

area of section required ' =39-3 square inches. 

B 

122:^ J 



I 

Center Line of Truss 



A 
POST U,L„ 

Fig. 3. 

The following trial section (Fig. 3) may be assumed: 

2 plates 22"XtB" =24.75 square inches 

4angles 3i"X3i"XA" = i4-6o 



39-35 



The plates are spaced 14 inches back to back. The 
moment of inertia about the AB axis will be: 



1000 



Plates = 2 ^ = 

Angles =4[4.o + (3-65 X io2)] = 1476 



Total = 2476 



.*• »' = \i^rT^ =7-95 inches. 



'39-35 



Art. 9.] DESIGN OF I^ERTICAL MAIN IVEB MEMBERS. I97 

It will not be necessary to make use of the radius of 
gyration about the CD axis (in this case 6.48 inches), since 
the length of the column to be inserted in the formula, 
when considering this radius, is much reduced by the 
cross-bracing at its upper end. The allowed intensity of 
stress for this shortened column, taken in connection with 
the smaller radius of gyration, should always be tested. 

In the present case the value of 7.95 inches for the AB 
radius checks the value assumed sufficiently near so that 
recalculation will not be necessary; the trial section will 
be adopted as it stands, using single 2^2 X t inch lattice bars. 

Member U3L3. 

Dead-load stress = —32,000 pounds 
*Live-load stress =+25,000 " 
Live-load stress = — 73,000 ' ' 

In the case of this member, the following specification 
relative to alternate stress becomes operative: 

§ jj. Members subject to alternate stresses of tension and 
compression shall- be proportioned to resist each kind of 
stress. Both of the stresses shall, however, be considered as 
increased by an amount equal to ^ of the least of the two 
stresses for determining the sectional areas by ihe usual 
allowed unit stresses. 

The equivalent live-load stress for UsLs then becomes 
(J dead + ^ tension live -|- compression live) load stress 
= — 109,000 pounds. 

Assuming a width of 15 inches, and the radius of gyra- 
tion .35 the width, the length of- the column being 27 feet 
(not 54 feet, since it is supported in two directions at the 
centre) furnishes a unit stress of 8500— (45 X 324) 75.25 = 

*See Chap. II, Art. 21, for the determination of this counter-stress. 



198 DET/tlLED DESIGN OF A RAILROAD BRIDGE. [Ch. VL 

5720 pounds per square inch. The required area will 
therefore be 

109,000 

= 1 0.0 square inches. 

5,720 ^ ^ 

A trial section of 2-15 inch channels, weighing 32.95 
pounds per linear foot, with a total area of section of 
19.38 square inches, will be found to furnish a radius of 
gyration of 5.67 inches, and to fulfil all the required con- 
ditions. The spacing back to back of channels will be 
made uniform with that of the other hangers and posts 
at 14 inches, and the latticing will be single, 2JXI inches. 

Member U5L5. 

Dead-load stress = — 7,000 pounds 
Live-load stress =-1-27,000 " 
Live-load stress = — 79,000 " 

The equivalent live -load stress is (JX 7000-1-^X27,000 
+ 79,000) = 104,100 pounds compression. 

The effective length of column may be assumed as one 
half of 63 feet, and since the stress in U5L5 is nearly the 
same as that in U3L3, the section belonging to the latter 
may be used for a trial; therefore r = 5.67 inches. The 
allowed intensity of stress then becomes: ^ = 8500 — 
(45X378)75.67=5500 pounds per square inch, and the 
area of cross-section required 104,100/5500 = 18.95 square 
inches. 

The trial section of 19.38 square inches may therefore be 
adopted as final, using as in the other cases a spacing of 
14 inches back to back of channels and single latticing 
2 J inches by Xf inch. 



Art. io.] 



DESIGN OF MAlhl IVEB MEMBERS. 



199 



Art. 10. — Design of Main and Counter "Web Members, 

Tension Web Members, U2U, U3M4, M4L5, t/sMg. 

The following specification applies to these tension web 
members : 

§ JO. The allowed unit stresses in tension for main 
diagonals and counters shall be: 

10,000 pounds per square inch for live loads. 
20,000 pounds per square inch for dead loads. 

Member U2L3. 

Dead-load stress = + 78,000 pounds 
Live-load stress = + 160,000 " 

The equivalent live -load stress is (J dead + live) load 
stress = + 199,000 pounds. 

The required area of cross-section will therefore be 



199,000 
10,000 



= 19.9 square inches 



and will be formed of 2 eye-bars 7 inches Xi^ inches = 
20.12 square inches. 

The following table shows the method of design for 
the other web members. 



Member. 


Dead-load 
Stress in 
Pounds. 


Live-load 
Stress in 
Pounds. 


Equivalent 

Live-load 

Stress 

in 

Pounds. 


Area 

Required , 

Square 

Inches. 


Section Adopted. 


Area of 

Final 
Section, 
Square 
Inches. 




+ 90,000 
+ 57,000 
+ 36,000 


+ 204,000 
+ 168,000 
+ 173,000 


+ 249,000 
+ 196,500 
-\- 191,000 


24.9 
19.65 
19. 1 


2bars-7"Xiif" 
2bars-7"XiT^" 
2bars-7"Xi|" 


25-38 
20. 12 
19.26 



2 00 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 

Counter Web Members, L2U3, L^M^, L^Me, and M4U5. 

The following specification applies to these members: 
§ 4j. The areas of counters shall be determined by taking 
the difference in areas due to the live- and dead-load stresses 
considered separately; the counters in any one panel must 
have a combined sectional area of at least j square inches, 
or else must be capable of carrying all the counter live load 
in that panel. 

Member L2U3. 

There is actually no counter stress in the panel L2L3, 
and the member L2U3 is only inserted to secure a margin 
of safety. In compliance with the specifications, there- 
fore, it must have a cross-section of at least 3 square inches. 
One adjustable rod if inches square having an area of 
3.06 square inches fulfils these requirements. 

Member LzM^. 

Dead-load stress = —90,000 pounds 
Live-load stress = -F 82,000 " 

The difference between the equivalent of the dead-load 
stress in terms of the live load and the live load will be 
— ^(90, 000) -1-82,000 =37,000 pounds tension. 

At 10,000 pounds per square inch a cross-sectional 
area of 3.7 square inches will be required. Two adjustable 
rods i^ inches square having an area of cross-section of 
4. 5 square inches will be used. 

The stress in the remaining counter members are not 
completely determined statically, and the sectional areas 
of those members should be based on the undiminished 
live-load stresses in them. 



Art. io.] DESIGN OF COUNTER IVEB MEMBERS. 201 

Member L^Mq. 

Dead-load stress = — 36,000 pounds 
Live-load stress = -|- 136,000 " 

This member should be designed according to the last 
condition of the specification, viz., to be capable of 
carrying all the counter live load in the panel ; the area 
required will therefore be 13.5 square inches, and will be 
composed of 2 adjustable bars 6 X 1 1 inches having an 
area of 13.5 square inches. 

Member M4U5. 

Live-load stress = -I- 39,000 pounds 
Dead -load stress = -h 62,000 " 

or 

Dead-load stress = — 66,000 pounds 
Live-load stress = +104,000 " 

This member will be designed according to the same 
condition as L^Mq. In this case all the counter live-load 
stress equals 104,000 pounds, and at 10,000 pounds per 
square inch the area required would be 10.4 square inches, 
which is supplied by two 6X1 inch eye-bars having an 
area of 12 square inches. 

This completes the design of the main members of the 
truss; the intermediate struts M3M4, MsMq, and the 
collision strut supporting the centre of the end post carry 
no definite stresses. Their form of section depends to a 
great extent upon their end connections, and they are 
usually built of light angles latticed together in box shape. 

In the present case all these struts will be built of four 
3^X3^X1 inch angles, having a combined area of 10 square 
inches laced together in a box form by four sets of single 
lacings 2jXA inches. The dimensions of the box forms 
will depend upon the connections which these struts make 
with other members. 



202 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VL 



Art. II. — Combined Stresses. 

§ jp. When any member is subjected to the action of both 
axial and bending stresses, as in the case of end posts of 
through bridges, or of chords carrying distributed floor loads, 
it must be proportioned so that the greatest fibre strength will 
not exceed the allowable limits of tension or compression on 
that member. If the fibre stress resulting from weight only, 
of any member exceeds lo per cent, of the allowed unit stress 
on such member, such excess must be considered in proportion- 
ing the areas. 

In accordance with the above specification the chord 
members should be tested with reference to the bending- 
stresses induced by their own weight. The condition of 
the ends or of the supported sections of these sections 
considered as beams is largely a matter of conjecture; they 
may be considered fixed at their ends or assumed entirely 
free to turn, or their condition may be assumed half way 
between. It is probably reasonable, at least, to assume 
these members fixed at their ends or supported sections, 
since the friction on the pins will tend to produce that 
condition. 

It will be sufficient for present purposes to test one 
upper and one lower chord member for combined stresses. 
The method here adopted for finding the bending stress in 
the truss members, although always yielding safe results, 
is not exact. Strictly speaking, the bending effect of the 
axial stresses should be recognized.* 

Member U^Uq. 

The cross-sectional area of this number is 99.09 square 
inches, but allowing 20 per cent, additional for rivets, 

* For the exact theory, see Burr's "Resistance of Materials," edition 1903, 
p. 181. 



Art. II.] COMBINED STRESSES. 203 

latticing, and other details, the weight per inch of length 
will be 33.8 pounds. The length between centres of end 
pins is 350 inches ; therefore the maximum bending moment 
M, assuming the member to have fixed ends, will be at 
the centre, and equal to 518,000 inch-pounds. 

The moment of inertia / of the section is 12,254; there- 
fore k, the intensity of stress in the extreme fibres, is, by 
the following formula, 

, Md 518,000X14 , . , 

k = ~j-=- = 592 pounds per square mch. 

This is less than 10 per cent, of the allowed compressive 
intensity in the member, and it may therefore be neglected. 

Member LsLq. 

The dimensions of one eye-bar are 8 inches deep X 1 1 
inches thick X 3 50 inches long ; the weight per inch of length 
will be 4^ pounds, and the maximum bending moment 
at the centre, assuming fixed ends, will be 32,500 inch- 
pounds. The moment of inertia is 80; therefore the 
stress in the extreme fibres at the centre will be 

32,500X4 ^ J -1 

- — "h =1625 pounds per square inch. 

The permissible axial unit stress (see page 186) was 15,750 
pounds per square inch, and the bending stress is so slightly 
in excess of 10 per cent, that it may be safely neglected. 

There still remains to be treated the bending in the 
end post caused by the overturning effect of the wind; 
the maximum bending will occur at the connection of the 
curved bracket of the portal to the end post and at the 
foot of the end post, since that member may be considered 
fixed at the lower end, and may be treated precisely as in 



204 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VL 

the case of a post supporting a roof-truss by the aid of 
brackets (Chap. I, Art. i8). .A point of contrafiexure 
midway between the foot of the end post and the curved 
bracket may therefore be assumed to exist. 

If it be supposed that the wind loads of the upper chord 
are carried equally to both end posts, the maximum bend- 
ing moment becomes \ the wind reaction by \ the distance 
along the post measured from its foot to the bracket con- 
nection; in this case 14,875 pounds X 140 inches = 2,082,500 
inch-pounds. The moment of inertia about the axis of 
bending is about 12,500, and the distance from the neutral 
axis to the extreme fibre of the upper angles is 15 inches, 
it being remembered that the bending takes place in the 
plane at right angles to the vertical plane through the 
axis of the truss. The stress in the extreme fibre will 
then be 

2,082,500X15 , . , 
=2500 pounds per square mch. 

At the same time the stress in the web plates, which con- 
stitute the greater portion of the cross-section of the post, 
will be (2, 082, 500X II. 5)/i2, 500 = 1920 pounds per square 
inch, due to this same bending. 

This stress is a wind stress, and is therefore subject to 
§ 36 of the specifications. The average- stress in the end 
post due to dead and live loads is (3 7 1,000-1-5 2 8,000) 797.2 5 
= 9250 pounds, and one quarter of this is 2310 potinds; 
therefore the bending due to wind may be safely neglected. 
As a further margin of safety it should be noted that the 
point of greatest bending due to wind does not coincide 
with the point of greatest bending due to the simple com- 
pression in the member. 



Art. 12.] DESIGN OF STRINGERS AND FLOOR-BEAMS. 205 



Art. 12. — Design of Stringers and Floor-beams. 

The following specifications apply to these members: 

§ JO. The allowed unit stress in tension, for the bottom 
flanges of riveted cross-girders {net section), shall he 10,000 
pounds per square inch, and for the bottom flanges of riveted 
longitudinal plate girders used as track stringers {net section), 
10,000 pounds per square inch. 

§ 40. In beams and plate girders the compression flanges 
shall be fnade of the same gross section as the tension flanges. 

§ 41. Riveted longitudinal girders shall have, preferably , a 
depth not less than one tenth the span. 

§ 42. Plate girders shall be proportioned upon the sup- 
position that the bending or chord stresses are resisted entirely 
by the upper and lower flanges, and that the shearing or web 
stresses are resisted entirely by the web plate; no part of the 
web plate shall be estimated as flange area. 

The distance between centres of gravity of the flange 
area will be considered as the effective depth of all girders. 

§ 4J. The webs of plate girders must be stiffened at 
intervals of about the depth of the girders, wherever the shearing 
stress per square inch exceeds the stress allowed by the follow- 
ing formula: 

Allowed shearing stress = 

12,000 



//2 

I+- 



3,000 

where H = ratio of depth of web to its thickness; but no web 
plates shall be less than | of an inch in thickness. 



The following specifications for riveting also apply to 
the design of these members: 

§ J7. The rivets in all members other than those of the 



2o6 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 

floor and lateral systems must he so spaced that the shearing 
stress per square inch shall not exceed gooo pounds, nor the 
pressure on the hearing surface {diameter ^thickness of the 
piece) of the rivet-hole exceed ij,ooo pounds per square inch. 

The rivets in all members of the floor system, including all 
hanger connections, must he so spaced that the shearing stresses 
and hearing pressures shall not exceed 80 per cent, of the ahove 
limits. 

In the case of field riveting {and for holts) the above- 
allowed shearing stresses and pressures shall be reduced one 
third. 

Rivets and bolts tmist not be used in direct tension. 



Stringers. 

The stringers are assumed 350 inches long and 32^ 
inches back to back of angles, the depth of the stringer 
being taken a little less than one tenth of the span to permit 
the required track elevation. The actual depth of the vi^eb 
plate will be 32 inches, but the distance back to back of 
angles is increased by \ inch, so that the web will not 
project between the angles and cause difficulty in riveting 
the cover-plates. It should be noted that shallow stringers 
deflect more than deeper ones, thus tending to produce 
greater direct tensile stresses on the end-connection rivets. 

The maximum bending moment due to live load will be 
found for that position of loading in which the centre of 
gravity of the load is situated as far on one side of the centre 
of the beam as the point of maximum bending is on the 
other side, and a wheel load will always be foimd over this 
point of maximum bending. In the present case this 
point is found 14.2 feet from the left end of the stringer 
under the second driver of the locomotive, as shown by 
Fig. 4. 



Art. 12.] DESIGN Of- STRINGERS AND FLOOR-BEAMS. 207 

The reaction at the left end of the stringer for this 
position of the loading is 43,800 pounds; the maximum 
bending moment is then found as follows: 

M = 43,800 X 14.2 — 10,000 X 13 — 20,000 X5 =392,000 ft. -lbs. 

= 4,704,000 in. -lbs. 

The dead-load bending moment at the same point may 
be taken equal to the maximum dead-load bending moment 
at the centre, and is found as follows : 

The track is assumed to weigh 400 pounds per linear 
foot, and both stringers and bracing 375 pounds per linear 




Fig. 4- 

foot. The weight per linear inch per stringer will then 
be 32.29 pounds, and the maximum moment 

32.29 X (350)2 

5-^^- — = 494,000 mch-pounds. 

o 

The effective depth of the girder according to the speci- 
fications may be assumed at 30 inches, and since the allowed 
intensity of stress is 10,000 pounds per square inch, the 
required net section will be 

4,704,000-1-494,000 . , 

— = 17-3 square mches. 

30 X 10,000 

This flange or chord area will be afforded by two angles 
6X6 XA inches, having a gross section of 13 square inches 
= 10.7 square inches net, and one cover-plate 14 X A inches, 



2o8 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VL 

having a gross section of 7.9 square inches = 6.8 square 
inches net. Total 17.5 square inches net. 

Length of Cover-plate. 

The length of the cover-plate will be determined by- 
assuming that the bending moment follows a parabolic 
. law. If L represents the total length of the stringer, / the 
length of the cover-plate, A the total section of the flange, 
and a the total area of all the cover-plates counting from 
the outside and including the area of the one under con- 
sideration, then 



/=Lv,^. 



Inserting the requisite quantities in the formula, 



,6.8 „ ^ 
2Q.2V =18.^ feet. 

At least 6 inches should be added to each end of the com- 
puted length of a cover-plate in which to place additional 
rivets, so that the plate may actually develop its full 
strength at the section where it is needed. The length 
of the cover-plate in this case will therefore be 19.3 feet. 

Weh Plate. 

The maximum shear for both dead and live loads occurs 
at the end of the stringer. The live-load shear with the 
first driver at this point is 62,500 pounds. The dead- 
load shear is 5700 pounds, thus making the total shear 
68,200 pounds. The working shear in the web plate may 
be taken at 5000 pounds per square inch of gross section. 
The required cross-section will then be 13.6 square inches, 



Art. 12.] DESIGN OF STRINGERS ^ND FLOOR-BEAMS. 209 

and the web plate taken will be 32XA" = i4.o square 
inches. According to the specifications, if the shearing stress 
exceeds 12, 000 /[i + (//^/^ooo)], intermediate stiffeners will 
be required. In the present case this formula has a 
value of i2,ooo/[i + (73.22/3000)] =4310 pounds per square 
inch; whereas the actual intensity of shear in the stringer 
is 5000 pounds per square inch at the ends, with a constantly- 
decreasing value towards the centre of the stringer; no 
intermediate stiffeners will therefore be required. The 
end stiffeners cannot be subjected to a rational analysis; 
in the present case they will be composed of two 6X6XA 
inch angles, being chosen of such dimensions as to permit 
of making simple connections to the floor-beams. They 
are sometimes designed by requiring that their normal 
section must carry the end reaction with a certain specified 
unit stress (as 7000 pounds per square inch). In this 
case the normal section is 10.12 inches, and it is ample 
for the purpose. 

Riveting. 

The following rule will determine the pitch in the flanges: 
of a girder, viz., the pitch of the riveting at any section 
of the flanges will be the quotient obtained by dividing 
the product of the vertical distance between the rows of 
riveting in the two flanges and the allowed stress in one 
rivet either for shear or bearing, by the maxim.um shear at 
the section. This rule is readily demonstrated by taking 
the difference between the bending moments at two adjacent 
rivets. 

In the present case the pitch at the ends of the stringer 
will be 

30 X (5740X80%) ^ 
68,200 



,2IO 



DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 



At lo feet from the ends the dead-load shear will be 
1925 pounds, and the maximum live-load shear, with the 
(iirst driver over the section, will be 31,300 pounds. 

The pitch at this point will therefore be 



30 X (5740X80%) 



= 4.15 inches. 



31,300-^-1925 

The flange rivets have thus far been considered to carry 
the direct horizontal stress only, but those in the upper 
flange must also carry as a vertical load the weight of the 
-driving-wheels as they pass. One driver weighing 20,000 




Fig. s. 

pounds may be supposed to be distributed over three ties, 
or over a distance of 3 feet, assuming 6-inch spaces between 
ties 8 inches wide. The greatest effect in increasing the 
pitch will be found at the centre of the girder. At this 
point 9 rivets would approximately be required in a dis- 
tance of 3 feet, if the longitudinal stress alone existed; 
.and 20,000/4592=4.4 rivets for the vertical load alone 



Art. 12.] DESIGN OF STRINGERS AND FLOOR-BEAMS. 2U 

The prop er numb er of rivets for this distance of 3 feet 
is then v'92 + 4.42 = 10. It is seen that the effect of this 
vertical loading in increasing the number of rivets is very- 
small. The spacing finally adopted is shown in Fig. 5; 
it varies from if inches at the ends to 3 inches at the 
centre, and more than provides for vertical load effect. 

Rivets in Cover-plates. 

The stress carried by the cover-plate will be 6.8 square 
inches X 10,000 pounds = 68,000 pounds. The permissible 
stress in the rivets will be determined for single shear, 
at 80 per cent, of 5410 pounds =4328 pounds per rivet. 
Sixteen rivets will be required. They will be placed in 
four rows with 3 inches pitch for a distance of 15 inches 
from the ends of the plate, and 6 inches pitch for the 
remaining distance. 

Rivets on End Connection. 

The rivets fastening the stringers to the floor-beam are 
field rivets, and therefore the allowed intensities of stress. 
in them must be reduced one third. 

Two separate conditions affect the number of rivets at 
this end connection; first, the single shear on these rivets- 
when one stringer has its maximum reaction, and secondly, 
the bearing on the web of the floor-beam, when the com- 
bined reactions at the two stringer ends resting on the 
floor-beam have their maximum value. For the first case, 
each rivet in single shear will carry 5410X. 8x1=2885 
pounds, and with a maximum end reaction of 68,200 potinds, 
there will be required 24 rivets; in the second case, assum- 
ing the thickness of the floor-beam web plate at A inch, 
the bearing capacity of one rivet will be 738oX.8xf = 
3930 pounds; the simultaneous maximum combined reac- 
tions of the two stringers will be, for dead load i4,oo<> 



212 



DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 



povinds, and for live load 84,000 pounds (see design of 
floor -beam for these reactions), or a total of 98,000 pounds; 
there will therefore be required 25 rivets. Thirteen rivets 
will however actually be placed in each 6x6 inch angle, 
making 26 rivets at this connection. 

The bill of material with the weights for one stringer, 
and one half the cross bracing, using that indicated in 
Fig. 5, for which no actual design will be made, may be 
written as follows: 

Lbs. per Ft. 

1 web plate 32X^" 29.1 feet long @ 50.58= 1470 lbs. 

4angles 6X6XA" " " " ©21.9 =2550 

2 cover-plates I4X^" 19-25 " " @ 26.78=1060 

4 end stiffeners 6X6X^" 2.6 " " ©17.2 = 178 

4 filling plates 6X^" 1.6 " " ©11.48= 74 

2^ bracing angles 3^X3|X:^" 9-5 " " @ 2.1 = 50 

About 450 pairs of |" rivet-heads @ 22.2 lbs. per 100 = 200 



Total = 5582 

The actual weight per foot of stringer is thus seen to be 
384 pounds, or 9 pounds per foot more than was assumed. 
This difference is so small as to require no correction in 
the design, but it is better to have the actual weight under 
rather than over that assumed. 

Intermediate Floor-beams. 

The maximum live-load floor-beam reaction occurs 
when wheel 5 is placed directly over the floor-beam. Wheels 
I to 4 will then be en one adjoining stringer and wheels 5 to 
9 on the other, making the resulting reaction 84,000 pounds. 

Assuming the dead weight of the .floor-beam to be 300 
pounds per linear foot, the dead-load reaction will become: 

Dead weight of one stringer = 11,300 pounds. 
Dead weight of ^ floor-beam = 2,700 



14,000 pounds. 



Art. 12.] DESIGN OF STRINGERS AND FLOOR-BEAMS. 213 

The sum of the dead- and hve-load reactions will then 
be 98,000 pounds. 

The stringers are spaced 7 feet 6 inches apart, and 
since the distance between trusses is 18 feet, the stringer 
is distant 63 inches from the centre line of the truss. 
Therefore the maximum bending of the floor -beam will be 
98,000X63=6,170,000 inch-pounds. Assuming an effect- 
ive depth of 42 inches, the flange area required will be 
6,170,000/42X10,000 = 14.69 square inches. This section 
will be composed of 2 angles, 6X4Xi inches = 9.6 square 
inches gross = 8.6 square inches net; i cover-plate 14 X J 
inches = 7.0 square inches gross = 6.0 square inches net. 
Total, 14.6 square inches net. 

The cover -plate should have the following length: 



/ = i8v — 2 = 1 1 ■ S feet ; 
^i 14-6 -^ 



owing to the details at the ends of the floor-beam the upper 
cover-plate will have a length of 12 feet 7 inches and the 
lower a length of 14 feet 9 inches. 

On account of the large number of rivets made necessary 
at the end connection of the floor-beam, and the consequent 
weakening of the plate, the assumed intensity of shearing 
stress for the web has been taken at 4000 pounds per 
square inch of gross section. The area required will there- 
fore be 98,000/4000 = 24.5 square inches, and a plate 43 X A 
inches = 24.2 square inches will fulfil the requirements. 
No intermediate stiffeners will be required, as the end 
connections of the stringers render end stiffeners unneces- 
sary. The distance from back to back of the flange angles 
will be 43 J inches. At the connection of the floor-beam 
to the post, a considerable portion of the web plate must 
be removed in order to leave room for the heads of the 
lower-chord eye-bars. It is therefore customary to cut 



214 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 

the web plate at some distance from the end, and to splice 
to it another plate of irregular shape having an equal cross- 
sectional area. The general detail drawing, Plate II, 
illustrates the method adopted in the present case. The 
central portion of the web is lo feet 12 inches long and 
spliced to it at each end by means of two A-inch plates, 
placed over J-inch fillers, is another plate designated on 
the figure as "web 24 inches X A inch X 6 feet 3 inches 
long." To this end web plate are fastened the end stiff eners 
formed of two 6X3jXAinch angles. The end stiff ener 
must be divided into an upper and lower part in order to 
allow the upper flange of the floor-beam to abut against 
the post. The lower portion of the end web plate will 
be cut away with rounded edges, to the form required by 
the disposition of the eye -bars, all floor-beams being made 
alike to conform to that post in which the greatest clear- 
ance is required. This edge will be protected by two angles 
3^X3^X1 inches, and 3 feet 6 inches long. The lower 
flange of the floor -beam will be connected to the foot of 
the post by a flat plate riveted to both, to provide against 
swaying, and to form part of the lower lateral wind system. 

Riveting. 

Instead of employing the rule used in finding the pitch 
of the rivets in the flanges of the stringers, it will be better 
to divide the amount of stress in the flanges of these beams 
between the post connection and the stringer connection, 
by the allowed stress capacity of one rivet; the quotient 
will determine the nximber of rivets between these two 
points. For example: 

146,900 

— 0,^0 =25 nvets. 
7380 X. 8 ^ 



Art. 12.] DESIGN OF STRINGERS AND FLOOR-BEAMS. 215 

These are easily provided for in the lower flange, but a pair 
of supplementary angles 6 X 3^ Xt^ inches and 3 feet i \ inches 
long must be fastened to the upper flange at its ends in 
order to provide for the requisite rivets to transfer the 
stress from the web plate to the flanges. The rivet design 
of this end connection is not very precise. The rivets 
piercing the spliced central portion of the web plate of 
the floor-beam really aid in carrying the stress through 
the splice-plate to the flanges. It is essential that all 
parts at this point shall be firmly and rigidly tied together. 
The pitch in the central portion of the beam will be 6 inches. 
The number of rivets necessary on each side of the 
splice will be 

98,000 



7380X.8 



17, 



and they will be placed in two rows, as shown in Plate II. 

At least the same number of rivets will be necessary to 
fasten the end stifiieners to the web plate; but since 17 
rivets cannot be placed in one row, the two rows shown 
on the plate will be inserted, even though all those there 
shown are not necessary. The resulting surplus may be 
considered as compensating for the tendency to pull on 
the rivet-heads, due to the deflection of the loaded beam. 

The end stiffeners are connected to the post by field 
rivets. The thickness of the web of the post is three 
eighths inch, but this thickness will be increased by a 
diaphragm which will later be inserted between the channel 
webs, so that the single shear of the rivets determines their 
number. 

The rivets required will therefore be [98,000/(5410 X 

.8Xf)] = 34. 

The rivets in the cover-plate will be pitched 3 inches 
for a distance of 1 5 inches from the ends of the cover-plate, 
and 6 inches for the remaining distance. 



2l6 



DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VL 



A small angle bracket consisting of a 3JX3JXI inch 
angle, i foot J inch long, will be riveted to the floor-beam 
to form a seat for the ends of the stringer during erection. 
This bracket is used for temporary convenience only, and 
is not supposed to carry any stress after erection. 

It is sometimes customary, if web plates have no splices, 
to take one sixth [or sometimes, one eighth] of the web 
section as acting in either flange. If no rivet -holes were 
punched for the stiffeners, this method would be allowable. 
But such rivet -holes frequently take out considerable 
metal, and as the tension side of the plate only is affected, 
one sixth of the remaining metal ceases to be a proper pro- 
portion. On the whole, therefore, it is better to neglect 
the bending resistance of the web, and allow it to balance, 
so far as it may, the effect of the rivet-holes being out of 
the centre of gravity of the flange angles. 

The bill of material and the weight of the floor-beam 
is given in the following table: 



Length. 



Weight per 
Foot. 



Total 
Weight, 
Pounds. 



1 web plate 43X1^" 

2 ' ' plates 24X ^" 

2 flange angles 6X4XI" 

2 " " ' ' 

1 cover-plate 14X i" 

1 " " " 

4 end stiffening angles 6X3iXA" 

4 " " " " 

4 " fillers 6Xt^" 

4 " finishing angles 3^X3^X1" 

4 splice-plates 24X1V" 

4 filling plates 30 X |" 

4 upper flange, end angles 6X3^X^" 

4 " " fillers 6Xi" 

2 " " " 3iXA" 

4 bracket angles 3^X3iXf" 

340 pairs of rivet-heads at 22.2 lbs. per 100. 



12 10 

6' 3" 

16' 10" 

14' 9" 



12' 

14' 
i' 
2' 

2' 
3' 
3' 
3' 
3' 



7" 
9" 
8|" 

7" 
4" 
6" 
6" 

ir 

ir 



02. 24 

45-9° 
16.2 
16. 2 
23.8 
23.8 
I3-S 
13-5 
8-93 
8-5 
35-7° 
51.00 

13 -S 
10. 20 
6.70 
8-5 



1040 
573 
545 
478 
300 

350 

90 

140 

83 

119 

500 

660 

163 

86 

IS 

35 

150 

5327 



Art. 13.] SPECIFICATIONS FOR LATERAL AND IVIND BRACING. 217 

The assumed weight is 5400 pounds, i.e., greater than 
the actual, as it should be. 

Art. 13. — Specifications for Lateral and Wind Bracing. 

The following specifications apply to the bracing: 

§ JO. Allowed unit stress, in tension, for longitudinal, 
lateral, and sway bracing for wind and live-load stresses = 
18,000 pounds per square inch. 

§ ji. Angles subject to direct tension must be connected 
by both legs, or the section of one leg only will be considered 
as effective. 

§ jj. Allowed unit stress, in compression, for lateral 
struts and rigid bracing =P = ij,ooo — yo l/r for wind stresses, 
where P, I, and r have the same meaning as previously . No 
compression member, however, sliall have a Jength exceeding 
12^ times its least radius of gyration. Lateral struts, with 
adjustable bracing, will be proportioned by the above formula 
to resist the maximum due either to the wind and load or to 
an assumed initial stress of 10,000 pounds per square inch 
on all the rods attached to them. 

§ 35. Members subject to alternate stresses of tension and 
compression shall be proportioned to resist each kind of 
stress. Both of the stresses shall however be considered as 
increased by an amount equal to eight tenths of the least of 
the two stresses, for determining the sectional area by the 
above allowed unit stresses. 

§ J/. The rivets in the lateral and sway bracing will be 
allowed jo per cent, increase over usual allowed unit stresses. 

§ p4. Where rods are used in the lateral, longitudinal, or 
sway bracing, they shall be square bars, but in no case shall 
they have a less area than one square inch. Rods with bent 
eyes must not be used. 

§ p5. All through bridges shall have latticed portals, of 
approved design, at each end of the span, connected rigidly 



2l8 



DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 



to the end posts and top chords. They shall he as deep as the 
specified head room will allow. 

§ g6. Where the height of the trusses exceeds 2j feet, an 
approved system of overhead diagonal bracings shall he attached 
to each post, and to the top lateral struts. 

§ p/. All hars and rods in the web, lateral, longitudinal, or 
sway systems must he securely clamped at their intersections 
to prevent sagging and rattling. 



Art. 14. — Design of Upper Longitudinal Wind Bracing. 

The tension members will be formed of angles connected 
by both legs, so that the full area will become available. 
The following table shows clearly the method of design, 
the allowable unit stress being 18,000 pounds per square inch.. 











Gross 










Area 




Area 


Net Area 




Member. 


Stress in 
Pounds. 


Required, 
Square 


Section Adopted. 


of 
Section. 


of Section 
in Square 


Rivets 
Required 






Inches. 




in Sq. Ins. 


Inches. 


at End. 


u,u. 


+ 42,500 


2.36 


i-3iX3^XA"angle 


2.88 


2.50 


6 


u,u. 


+ 33,000 


1.84 


i-"X"Xf" " 


2.50 


2.17 


5 


u^u. 


+ 23,700 


1-33 


i-"X"X" 


2.50 


2.17 


4 


u,u. 


+ 14,300 


0.80 


i-"X"XA" " 


2 .09 


1. 81 


2 


u,u. 


+ 4,700 


0.28 


i-"X"X" 


2.09 


1. 81 


I 



These members are connected to the upper chord by 
f-inch gusset-plates. The bearing capacity of one machine- 
driven rivet in this thickness of plate will therefore be 
4920 X 150 per cent. = 7380 pounds, and this quantity deter- 
mines the number of rivets required at the end connections 
as given in column 7, although all the connections will be 
made uniform with six rivets. The compression members 
at right angles to the axis of the bridge are not designed 
by an}^ exact theory. They will be formed either of cross 
frames or struts, as indicated on the stress sheet, and they 
are more than ample in section to sustain the direct com- 
pressive stresses there indicated; in fact, since their main 



Art. 14.] UPPER LONGITUDINAL IVIND BRACING. 219 

duty is to stiffen the structure and to cause the two trusses 
to act as a whole, they are made as rigid as the judgment 
of the engineer requires. The portal bracing is designed 
in a somewhat similar way, although tested by computa- 
tions in its various parts. 

The struts will be placed at those points, t/4 and Uq 
where the upper chord is continued in a straight line, and 
where there are no inclined web members. They will 
be composed of four angles, each Z^X^^Xh inches, built 
in I form with double lattice bars 2|X| inches, and con- 
nected to two gusset-plates, one lying on top of the cover- 
plate of the upper chord and the other fastened to the 
bottom angle of the upper chord. 

The cross frames will be placed at all the other panel 
points. They will be formed of an upper strut, like that 
just described, and a lower strut placed at the upper clear- 
ance limit and formed of four angles each zhXzhX^ inches, 
latticed together in I form by bars 2-| XA inches. The lower 
strut will be fastened to the post by means of a A-inch 
gusset-plate lying between two Z'hX^Xi^ inch angles, 
two feet long, the latter being riveted to the centre of the 
posts by twelve rivets. Between the two I struts will then 
be placed a compound lattice-work composed of single 
3^X31X1% inch angles tied together at intersections by 
small square plates with five rivets at each crossing and 
fastened to the struts by tfe-inch gusset-plates, the details 
of which are shown in Plate II. The cross frames are really 
the intermediate transverse bracing, the theory of whose 
design would be determinate if it were not the duty of the 
upper horizontal lateral bracing to transfer all the wind 
load to the upper ends of the end posts. It is possible to 
design the cross frames by ignoring this upper wind bracing. 
Their main function, however, is to stiffen the bridge under 
rapidly moving train loads, rather than to give additional 
stability against wind loads. 



220 



DET/IILED DESIGN OF A RAILROAD BRIDGE. [Ch. VL 



Art. 15. — Design of Portal Bracing. 

The portal bracing will be made as rigid as possible and 
will take the form of a latticed girder placed in the plane 
of the pins of the end posts, as shown in outline on the 
stress sheet. The upper and lower flanges will be made 
alike of two 6X4XI inch angles with a 14 X finch web 
plate. The two flanges will be tied together by a com- 
pound latticing composed -of single 5 X 3^ X f inch angles. 
The upper flange will be fastened to the cover-plate of the 
top chord, U1U2, by means of a bent gusset -plate, as shown 
in Plate II; the lower flange will be fastened to the side 
plate of the end post in a manner similar to that of the cross 
frames. The lower flange will, moreover, be strengthened 




29700 lbs. 



76500 lb3. 



M700 lbs. Lt 



Fig. 6. 



at each end connection by means of a plate and angle bracket, 
composed of a |-inch plate and siXsJXf inch angles, as 
shown in Plate II. The outer end of the bracket will be 
supported from the latticing of the portal frame by a 
single 3^X2^X1 inch angle. 

The wind force acting at the upper end of the end post 
is 29,700 pounds. The length of the end post is 46.3 feet 



Art. i6.1 DESIGN OF LOIVER IVIND BRACING 221 

and the distance between trusses 18 feet. The increase of 
load on one truss, which acts as if hung at the connection 
point of the lower flange of the portal, will therefore be 
76,500 pounds. The multiple bracing employed produces 
ambiguity in the stresses of the portal members. If the 
assumption be made, as shown in the accompanying figure, 
that the multiple bracing is replaced by a single tension-bar, 
and that the wind reaction is entirely resisted at the foot 
of the leeward post, the stress in the lower flange will be: 

— ^—7 = 82,200 potmds compression. The allowed 

intensity of stress is P = 13,000— 7o//r, in which I = 216 

inches and r may be taken approximately as 3 inches. 

Then P = 8ooo poiinds per square inch, approximately, 

82,000 
and the sectional area required will be -^ =10.4 square 

inches. 

The actual area will be: 

2 6X4X1 inch angles = 7.22 square inches. 
I 14X1 inch plate =5 25 



12.47 



The section adopted is therefore sufficient. 

The stress in the upper flange may be taken to be equal 
to the wind reaction transmitted from one truss to the 
other; its section, although so taken, need not be as great 
as that of the lower flange. 

Art. 16. — Design of Lower Longitudinal "Wind Bracing. 

Although the inclined web members in this horizontal 
truss may suffer apparently reversal of stress, it must be 
remembered that the two systems of bracing provide for 
opposite directions of the wind without such reversal. 



DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VL 



The specifications in regard to reversal of stresses do not 
therefore become operative. If it be assumed that the 
inclined web members act in compression, they must be 
designed with the prescribed unit compressive stresses. 
As they are fastened at two points to the stringers and 
at the centre to each other, it may be proper also to assume 
the effective length of a member to be the distance from 
its end to the point of intersection with the centre line of 
the stringer, i.e., in this case about 120 inches. These 
members will be formed of two unequal legged angles, 
with the longer legs vertical and riveted back to back. 
The following table shows the sections adopted : 



Mem- 
ber. 


Stress in 
Pounds, 
Compres- 
sion. 


Radius of 
Gyration 
in Inches, 

Final 
Section. 


Allowed 

Unit 

Stress 

in Pounds 

per 
Square In. 


Area 

Required 

in Square 

Inches. 


Section 
Adopted. 
2 AngUs. 


Actual 

Section 

in Square 

Inches. 


Number of 

Rivets 

Required 

at End 

Connection. 


L,L, 
L,L, 
L,L, 
L,L, 
L,L, 
L,L, 


71,600 
59,200 
47,700 
36,900 
26,800 
17.500 


1.56 
1.56 
1.56 
1.56 
1.23 
1.23 


7610 
7610 
7610 
7610 
6170 
6170 


9.4 
7,8 
6.3 
4-85 

4-35 
2.84 


SX4XA" 

" xr 

" Xf" 

" Xf" 

4X4Xf" 


9-52 
8.58 
6.46 
6.46 

5-7 
S'7 


14 
II 
10 

7 
6 

4 



It will be seen that the sections adopted for the panels 
near the centre of span are greater than actually needed; 
but since a portion of the duty of these members is to 
stiffen the structure londer rapidly moving loads their 
section may properly be greater than required by the 
computed stresses. 

The connection of these members to the posts is made by 
a fiat plate \ inch thick, which in turn is fastened to the 
foot of the post by 3|-X3iX finch angles, as shown in 
Plate II. The amount of stress which these angles must 
carry to the plate is the difference of wind stresses found 
in the two adjacent lower chord panels which has a maxi- 
mum value of 222,000—122,000 = 100,000 pounds at the 
foot of UiLi. 



Art. 17.] DESIGN OF BEDPLATES AND PEDESTALS. 223 

The Specifications for wind-bracing riveting show the 

single shearing capacity of one rivet to be 5410X150 per 

cent.X3 = 54io pounds and the bearing capacity in a 

f-inch thickness of metal to be 4920X150 per cent.X-| = 

4920 pounds. At the foot of t/iLi there will therefore be 

100,000 . . , . . , 

required = 21 rivets ; since the two angles mside 

^ 4920 ^^ 

the post do not afford space for all these rivets, a U-shaped 
plate will be riveted within the post, which will permit 
placing the required number of rivets. At the foot of 
U2L2, the maximum stresses carried by the plate will be 
78,000 pounds, and will require 16 rivets. At the foot of 
UaLs the stress will be 56,000 pounds, and 12 rivets are 
required. Plate II shows the disposition of these rivets 
in the angles. The number of rivets connecting the ends 
of the lateral bracing to the plate, as given in column 8 
of the above table, was found in the same way as for the 
upper chord. 

The lower flanges of the floor-beams act as struts in the 
lower lateral system. The ^-inch plate which is used as a 
connection at the foot of the posts is sufficiently extended 
to be firmly riveted to the lower flange of the beam. The 
number of rivets necessary may be determined by assum- 
ing that the plate transmits to the floor-beam that com- 
ponent of the stress in the inclined lateral members which 
acts in a direction at right angles to the axis of the chords. 

Art. 17. — Design of Bedplates, Friction-rollers, and Pedestals. 

§ 100. All bedplates must be of such dimensions that the 
greatest pressure upon the masonry shall not exceed 2jo 
pounds per square inch. 

^ loi. All bridges over 75 feet span shall have at one end 
nests of turned friction-rollers running between planed sur- 
faces. These rollers shall not be less than 2^ inches diameter 



2 24 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 

for spans loo feet or less, and for greater spans this diameter 
shall he increased in proportion of i inch for loo feet addi- 
tional. 

The rollers shall he so proportioned that the pressure 
per linear inch shall not exceed the product of the diameter 
in inches by joo pounds. 

§ 104. Pedestals shall he made of riveted plates and 
angles. All hearing surfaces of the hase-plates and vertical 
wehs must be planed. The vertical webs must be secured to 
the web by angles having two rows of rivets in the vertical legs. 
No base-plate or web connecting angle shall be less in thick- 
ness than f inch. The vertical webs shall be of sufficient 
height, and must contain material and rivets enough to prac- 
tically distribute the loads over the hearings or rollers. 

Where the size of the pedestal permits the vertical wehs 
must be rigidly connected transversely. 

§ 10 j. All the bedplates and bearings under fixed and 
movable ends must be fox bolted to the masonry; for trusses, 
these bolts must not be less than i| inches diameter. 

The dead-load reaction at the base-plate is 325,000 
pounds and the live load 420,000 pounds, a total of 

745,000 pounds. The masonry bearing will be '- = 

2980 square inches. The diameter of the rollers according 

250 
to the specification should be at least 2'^+ = 5! mches, 

but since it is better practice to build segmental rollers the 
diameter of these rollers may be taken at 8 inches. Assum- 
ing then that eight rollers will furnish a proper form of 
distribution, the number of linear inches of length per 

roller required will be x~-, — '- — --^ = ^8.9 inches; the rockers 
^ 8X300X8 

will therefore be 8 in number, 8 inches high, 5 inches wide, 

and 4 feet 5 inches long, rolling on a base plate i^ inches 

thick, 4 feet 2 inches X 5 feet 6 inches in area. A ij-inch 



Art. 17.] DESIGN OF BEDPLATES AND PEDESTALS. 225 

plate, 50 inches X 4 feet 6 inches, placed over the rockers 
will form the seat or bedplate for the pedestal proper. 
Both base-plate and bedplate will have riveted to them, 
so placed as to be directly below the centre of the end 
post, a flat bar, 2\ inches X f inch X 4 feet 2 inches long, 
made to fit corresponding grooves turned in the rockers 
to prevent all lateral motion of the rockers or pedestal. 
The rollers will be fastened i^gidly together in their proper 
positions with |-inch clear spacing by a f-inch strap on 
each side 2 feet 3 inches long secured to the axis of each 
roller by a nut. 

The pedestal must distribute uniformly to the shoe 
or bedplate the vertical load carried by the pin at the foot 
of the end post.. For this reason it should be made deep 
enough to prevent appreciable deflection, but not so deep 
as to cause a tendency to buckle the side-plates. A com- 
pletely rational design for a pedestal is practically impos- 
sible. It is treated in some cases as a short plate girder 
uniformly supported along the bottom and carrying a 
concentrated load at its centre. It is more usual to select 
some tried form of pedestal and adapt it for use in accord- 
ance with good judgment ; this will be done in the present 
case. The side-plates of the pedestal supporting the end 
pin must first be considered. The diameter of this piii 
should therefore be known, and in this instance may b^ 
assumed to be 7 inches. With an allowable bearing pres- 
sure of 15,000 pounds per square inch, the total thickness 
of the side-plates for one half of one pedestal will be 
745,000/2 -^ (15,000 X7) =3.55 inches. If the main side- 
plates are fastened between two 6x6Xf inch angles, 
■f filling-plates will be required above the angles. In 
addition to these two f-inch plates there will then be 
needed two H-inch plates and one f-inch plate, giving a 
total thickness of 3! inches. An additional plate ^ inch 
thick, called a lap-plate, projecting upward within the 



2 26 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 

end post, will be fastened to the inside of these plates. 
The riveting of these connections is clearly shown in Plate II. 
It will be seen that a i^-inch cover-plate and two 3^X3^X1 
inch angles are added to the upper edges of the pedestal 
in order to make it conform in finished appearance to the 
end post. 

The pedestal at the fixed end of the truss will be exactly 
similar to the- one above described, but the rollers will be 
dispensed with and the difference in height will be provided 
for in the pedestal itself. 

The base-plate, as shown in the figure, will be rigidly 
secured to the masonry by i^-inch anchor-bolts 2 feet long. 
On the outer side of the rollers, a dust-guard of two angles 
and a plate built in Z shape will be connected to the anchor- 
bolts. 

Art, 18. — Design of End Floor-beam. 

The end floor-beam, which is fastened to the shoe-plate 
and which carries not only the end stringers, but also the 
short cantilever-brackets supporting the track from the 
abutment itself to the end stringer, still remains to be 
designed. 

The end shear of this floor-beam will be the live-load 
shear of a single stringer, 62,500 pounds, together with an 
estimated dead-load shear of 7500 pounds, or a total of 
70,000 pounds. The maximum moment will then be 
70,000 pounds X 63 inches = 4,440,000 inch-pounds. Assum- 
ing the effective depth of girder to be 54 inches, the required 
sectional area of one flange will be 8.15 square inches, and 
it may be composed of two angles 6 X 4 X i inch with a 
gross section of 9.6 square inches and a net section of 
8.6 square inches. 

The v/eb will be a 55! Xf inch plate with an area of 
20.9 square inches. Allowing a compressive intensity. 



Art. 19] DESIGN OF PINS AND JOINT DETAILS. 227 

of stress of 8000 pounds per square inch in the end stiffeners, 
the area required will be 8.75 square inches. Two 6X4X2^ 
inch angles having an area of 9.5 square inches will be used, 
with fillers. 

The end floor-beam rests on and is riveted to a hori- 
zontal ^-inch plate, which in turn is solidly united .to the 
shoe-plate. To prevent lateral deflection, the upper flange 
of this floor-beam is tied to the end post by means of a 
horizontal bent plate, and in addition a vertical triangular- 
shaped plate is fastened to the end stiffeners and to the 
shoe-plates, as shown clearly by Plate 11. The plane of 
the lower lateral bracing does not coincide with the plane 
of the shoe-plate, and to provide a proper connection, two 
6 X 4 X i- inch angles are riveted to the web of the floor-beam 
at the proper elevation. 

The end-stringer bracket is also shown in sutflcient de- 
tail in Plate IL The floor-beam is slotted in order to allow 
the tie-plate forming the cover-plate of the bracket to pass 
through it. This tie-plate is riveted to the upper flange 
of the adjacent intermediate stringer, to relieve as much 
as possible the direct stress of tension which may C3:ne 
upon the rivet-heads connecting the bracket to the floor- 
beam. 

Art. 19. — Design of Pins and Joint Details. 

§ jS". Pins shall he proportioned so that the shearing stress 
shall not exceed gooo pounds per square inch; nor the crush- 
ing stress on the projected area of the semi-intrados of any 
member {other than forged eye-bars, see § 80) connected to 
the pin be greater per square inch than ij,ooo pounds; nor 
the bending stress exceed 18,000 pounds, when the applied 
forces are considered as uniformly distributed over the middle 
half of the bearing of each member. 

§ y8. The lower chord shall be packed as narrow as possible. 



■2 2 8 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 

§ 80. The diameter of the pin shall not he less than f 
the largest dimension of any eye-bar attached to it* The 
several members attaching to the pin shall be so packed as to 
produce the least bending moment upon the pin, and all 
vacant spaces must be filled with wrought filling rings. 

■ § 88. Where necessary, pinholes shall he reinforced by 
■ plates, some of which must be of the full width of the member, 
so that the allowed pressure on the pins shall not be exceeded, 
arid so that the stresses shall be properly distributed over 
the full cross-section of the members. These reinforcing 
plates must contain enough rivets to transfer their proportion 
of the hearing pressure, and at least one plate on each side 
shall extend not less than 6 inches beyond the edge of the 
.bottom plates (§ Sy). 

§ 8y. The open sides of all compression members shall 
be stayed by batten plates at the ends. . . . The batten plates 
must be placed as near the ends as practicable, and shall 
have a length not less than the greatest width of the member, 
or i\ times its least width. 

§ 8g. Where the ends of compression members are forked 
to connect to the pins, the aggregate compressive strength of 
these forked ends must equal the compressive strength of 
the body of the members. 



In addition to these direct specifications the following 
may be noted: 
..; . % 60. In compression members, abutting joints with 

* The following analysis furnishes the foundation for this specification. If 
the width of an eye-bar be represented by w, and the depth by h (the thickness 
■of the head is taken the same as that of the bar), and if T be the permissible in 
, tensity of tensile stress, the total stress carried by the bar will ht w-h-T. 1 . 
>the diameter of the pin be represented by d and C be the permissible bearing 
intensity of stress on the pin, then d-w-C will represent the bearing allowed on 
the pin. The two quantities wh-T and d-w-C should be equal; if the ratio 
of" C/T be taken at 4/3, then whT=dwC and d=T,/^h. 



Art. 19.] DESIGN OF PINS y^ND JOINT DETAILS. 229 

planed faces must be siijjicieiitly spliced to maintain the 
parts accurately in contact against all tendencies to dis- 
placement. 

§ 61. In compression members, abutting joints with 
iintooled faces must be fully spliced, as no reliance will be 
placed on such abutting joints. ... 



The bending in pins and the aUowed bearing of the 
plates on pins are pecuHariy related; the larger the 
diameter of the pin, the narrower are the necessary bearing 
surfaces and consequently the smaller are the bending 
moments. If the diameter of the pin be decreased, the 
bearing surfaces must be increased in width. In that 
case the bending moments are increased and the pin 
diameter may have to be increased. It will usually be 
impossible to obtain an economical balance for these 
conflicting conditions throughout the truss without many 
different sizes of pin. It is therefore customary to fix upon 
one size of pin for the lower chord, and another smaller 
one for the upper chord, then test them for bending and 
design the bearing surfaces to correspond with the diameters 
so chosen. The packing or arranging of the eye-bars must 
be so chosen as to reduce the bending to a safe minimum. 

Specification § 80 at once fixes a minimum limit for 
the size of the lower chord pins and also of the pins at the 
upper extremity of the end post at 6 inches ; and the pins 
in the upper chord at 5^ inches. The sizes tentatively 
assumed will be 7 inches for the first set of pins and 6 inches 
for the latter. The bearings of the various members for 
these sizes of pins are then designed, and then all the 
members at any one point so packed as to produce the least 
practicable bending moment. The "play" or spacing 
allowed in packing eye -bar heads may sometimes be taken 



230 DETAILED DESIGN OF A RAILROAD BRIDGE. [Cii. VI. 

as small as h inch to allow for slight imperfections in 
manufacture, but it is usually taken as \ inch for each 
eye-bar head. This amount may be increased to meet 
the requirements of any special case. The clear distance 
between built members such as chords or posts is usually 
greater; it may be taken as high as i inch. It will be 
found necessary at some connecting points to cut away 
parts of the flanges of channels or angles in built-up members 
on account of the interference of inclined members. If 
this must be done, the remaining part of the member must 
always be tested by computation as to its strength, con- 
sidering it as a short piece in compression. In most cases 
the usual pin-bearing' plates are sufficient to replace the 
metal cut away. The following formula has been much 
used for the purpose of designing such forked portions 
of the ends of posts: 

P l_ 

in which t represents the total thickness of metal whose 
width is 6 ; P the total load on one jaw of the post (usually 
one half the total load carried by the post or cohimn) 
and / the distance from the centre of the pinhole to the 
last centre line of rivets in the body of the column back 
of the cut in the angle or in the flange of the channel. 
This formula is applicable to steel with ultimate tensile 
resistance running from 60,000 to 68,000 poimds per square 
inch. For higher steel, or for highway bridges, or for 
other structures where less margin of safety may be justi- 
fiable, the value of t may be made correspondingly less 
than given by eq. (i). 



Art. 19] DESIGN OF PINS AND JOINT DETAILS. 231 

Pin-plates in Posts. 

The posts at their lower extremities carry to the pins 
not only their maximum stresses as members of the truss, 
but also the compression due to their acting as supporting 
struts for the floor-beams. It is possible and it should be 
assumed that the maximum post stress and the greatest 
panel load occur together. In the present case this re- 
quires that the lower ends of the posts be reinforced with 
sufficient thickness of bearing plates to carry 98,000 pounds 
in addition to the stresses taken from the stress sheet 
(Plate I). In order that this floor-beam load may be dis- 
tributed uniformly to the pin by both jaws of the column, 
it is necessary that the two sides of the post be riveted 
firmly together to act as a whole ; otherwise the inside jaw 
will carry the greater portion of the load. This is usually 
accomplished by inserting between the two parts of the 
column, at the end of the floor-beam, a diaphragm which 
acts as a continuation of the web plate of the beam. This 
diaphragm will be composed of a |-inch plate riveted 
between four 4 X 3 X | inch angles. The method of fastening 
is clearly shown in Plate II. 

Member LiU\ — Lower End. 

This member needs to be fastened at its lower end 
only to increase the general stiffness of the structure, since 
it is simply a hanger to carry load to the upper panel point. 
The U-shaped plate to which the lower lateral bracing is 
attached furnishes inside pin-plates, but in addition two 
f-inch plates will be fastened to the outside extending 
upward to the bottom of the floor-beam, so that the latter 
may have support during erection. 



232 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 

Member LiUi — Upper End. 

A I -inch plate, bearing on a 7-inch pin, may carry 
105,000 pounds. The load to be carried at this point is 
117,000 pounds; therefore 1.12 inch total width of bearing 
will be required. As the webs of the channel are each 
0.375 inch thick, the thickness of pin-plate required would 
be 0.37 inch, but since the net section of the channels 
is reduced by the diameter of the pin, it will be advisable 
to use two |— inch pin -plates, especially as the vertical 
distance from the top of the pin to the upper extremity of 
the channels is but 6 inches. The net section through 
•this point should not be less than two thirds of the transverse 
net section, and the f-inch pin-plates chosen will satisfy 
the condition. 

Since the allowable stress in the main section is 6000 
pounds per square inch, each pin-plate may be supposed 
to carry 15x1x6000 = 56,300 pounds, and as the rivet 
bearing in a f-inch plate is 4920 pounds, twelve rivets will 
be required. 

Member U2L2 — Lower End. 

The sum of the dead and live loads carried at this point, 
together with the floor-beam load of 98,000 pounds, may 
produce a total load of 357,000 pounds, or 178,500 pounds 
on each jaw of the post resting on a 7-inch pin. A total 

thickness of '- — =1.7 inches will therefore be required. 

105,000 ^ ■ 

A |-inch inside, and a |-inch outside pin-plate, together 

with the A-inch main plate, give a thickness of 1.81 inches, 

and will prove satisfactory. 

Since the bearing value for one rivet in a A-inch plate 

is 7380 pounds, or less than the double shear value, the 

former quantity will determine the number of rivets. It 



Art. 19.] DESIGN OF PINS AND JOINT DETAILS. 233 

may be assumed that the total stresses carried by the 
different plates are proportional to their thickness; there- 
fore (f +-|) ^lA =69 per cent, of the total stress must be 
transferred by rivets from the main plate to the pin-plates. 
In this case 0.69 X 178,500 = 123,000 pounds; therefore at 
least seventeen rivets will be required; Plate II shows the 
arrangement of plates and rivets. 

Member U2L2 — Upper End. 

The 6-inch pin at this point will carry a bearing pressure 
of 90,000 pounds per inch width of plate, the load on each 
jaw of this post being 129,500 pounds. 1.44 inches thick- 
ness of bearing will be required, and this thickness will be 
afforded by the A-inch main plate, with a ^^-inch outside 
and a ^-inch inside pin-plate. The bearing value of one 
rivet in a i^-inch plate is 7380 pounds, and the rivets transfer 
63 per cent, of 129,500=81,500 pounds, to the pin-plates, 
hence the required number will be 11. Plate II shows 
the arrangement adopted. 

Member U3L3 — Lower End. 

The load carried by one jaw resting on the 7 -inch pin, 
including the floor-beam load, will be 101,500 pounds, 
and the thickness required will be ■roioo^=o-97 inch. A 
f-inch pin -plate in addition to the f -inch web of the channel 
will therefore suffice. The stress transferred to the pin- 
plate will be 63,500 pounds, and as the bearing value of one 
rivet in a |-inch plate is 4920 pounds, thirteen rivets will 
be required. 

Member UzL^ — Upper End. 

The thickness of plates for one jaw bearing on a 6-inch 
pin will be loMo =o-58 inch. A J-inch pin-plate will be 



2 34 DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VL 

riveted to the outside of the channel and the number of 
rivets required will be 6. 

It will not be necessary to repeat these operations in 
detail for the other posts of this truss, the method of 
design being precisely the same as those already given. 



Art. 20. — Design of Pin Connections of Upper Chord Members. 

As the joint of two adjoining upper chord members 
having different inclinations is always at the angle in the 
chord with its plane passing through the axis of the pin, 
it is usually made abutting and riveted so as to make 
the chord continuous. This makes the chord stiff er than 
if the joints were pin-bearing. It has become almost 
the invariable practice, however, to make the joint at 
the top of the end post an open pin-bearing one. Since 
all abutting faces will be planed, specification § 6o requires 
no splice-plates except those sufficient to prevent dis- 
placement. The only stress transferred from a pin to a 
chord member will then be the resultant, in the direction 
of the chord member, of the maximum stresses in the 
web members meeting at that point, i.e., the increment of 
stress. Sufficient pin-bearing area must be provided to 
carry this increment of stress. 

The joint for adjoining chord members whose axes are in 
the same straight line is never made at a pin, but at some 
distance from it, on that side away from the centre. This 
procedure is adopted for convenience in erection, which 
usually proceeds from the centre towards the ends. This 
joint is placed as near the pin as practicable, and in a chord 
20 inches deep, it is usual to have three rows of rivets in 
the splice-plate on each side of the joint; for splices in 
chords 24 to 30 inches deep four rows are usually inserted. 
At such a joint a batten plate on the bottom and a short 
cover-plate on top are used to strengthen the splice. 



Art. 20.] PIN CONNECTIONS OF UPPER CHORD MEMBERS. 235 

It has already been noted that the axes of the pins have 
been placed 14^ inches below the edge of the cover-plate, 
or 0.7 inch below the centre of gravity of the chord section. 
This tends in some degree to neutralize the flexure due 
to own weight by the counterfiexure introduced by the 
increment of chord stress. 

In the bridge under consideration no joint will be 
made at panel points Ui and Uq] the members U^UiU^ 
and U^U^Ut will each be made in single pieces about 59 
feet long, which is entirely practicable. The supporting 
struts which frame into these members at their centre 
points are connected by means of angles, as shown in 
Plate II. The joint for U2 is made 2 feet 4% inches away 
from the centre of U-2 and /o-inch splice-plates will be 
riveted on the inside of the side-plates of the chord, with 
four rows of rivets on each side of the joint. A J-inch 
filler is necessary to allow for the difference in thickness of 
plates in the two chord members. The ^-inch outside 
plate will be continued to cover the joint. 

The maximum increment of stress in U2U3 given by 
the web m.ember at U2 is 140,000 pounds, or 70,000 pounds 
for each jaw. The 6-inch pin can transfer this stress 
to the H-inch side-plate, but for increased stiffness and 
solidity a ^-inch plate with J-inch filler will be riveted on 
the outside of each jaw. As may be observed, all con- 
nections are made so strong that a member should fail 
rather in its body than at its end connections. 

Connection at U3. 

The maximum increment of stress for the member 
UzUi at panel Uz in the direction of the axis of U3U4 may 
be found graphically from the stresses in the web members 
to be 200,000 pounds. The thickness of each jaw bearing 
on a 6-inch pin must then be at least i.i inches. It will 



236 



DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. Vi- 



be seen in Fig. 7 that besides the nj-inch filler-plate and 
the |-inch lap-plate, a §-inch inside and a |—inch outside 
pin-plate have been used. This is more than the com- 
putations require, but not more than desirable for the 
stiffness of the continuous upper chord. 

Plates corresponding to those riveted to U3U4 are 
fastened to U2U3 at U3. The outside edges of the main 
side-plates of adjacent members should always be in line. 
In this truss the uniform distance from back to back of 




Fig. 7. 

these plates is 23 inches. The abutting pieces at U3 are 
shown in Fig. 7. 

The stress carried by each plate will be determined by 
multiplying its thickness in inches by the allowed bearing 
of a I -inch plate on a 6-inch pin, and this quantity deter- 
mines the number of rivets required in each plate. These 
rivets must be designed not only for single shear for each 
separate plate, but also for the combined bearing of the 
outside plates on the main inside web plate. Where field 
rivets are used their stress value is only two thirds that of 
shop rivets. 

If the thickness of pin-plates is large, considerable 



Art. 21] DESIGN OF DETAILS AT ENDS OF END POST. 237 

bending, indeterminate in amount, may take place in the 
rivets, and it is proper on this account to add more rivets 
than those actually required by computation. Moreover, 
since the cover-plate and top angles of the chord section 
receive their stress cumulatively, the pin-plates which 
transfer a part of the stress to them should be made long 
enough to enable this stress to be transferred directly, 
instead of first passing through the main web plate. 

Since there is thus a certain amount of indetermination 
in the transmission of stress in an abutting joint, such a 
joint is not designed strictly according to computations, 
but more or less according to experience tempered by the 
engineer's judgment. The design adopted must, however, 
always be tested by some method of analysis to determine 
its safety. 

Connection at Uri. 



The pin-plates at this panel point are designed in a 
manner similar to that followed for Us. The design 
adopted, Plate II, provides abundant material. 



Art. 21. — Design of Details at Ends of End Post. 

Lower Extreinity . 

The end connections of LqUi are not abutting, but 
pin-bearing. Hence all the stress in LqUi, 899,000 pounds, 
is transferred to the pin and from that to the pedestal plates. 

A i-inch plate bearing on a 7-inch pin can carry 7 X 
15,000 = 105,000 pounds; therefore a 4.28-inch thickness 
of pin-plate is required for each jaw. There will then be 
added to the main web plates ^ inch and 1% inch thick 
respectively, an outside |-inch lap-plate, a -|-inch plate 
below this, and a i^c-inch plate over a |-inch filler. On 
the inside, next to the side-plate, there will be a /B-inch. 



238 



DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 



plate, and then a i^-inch plate; this gives a bearing thick- 
ness of 4 1 inches. The following rivets will be required 
in single shear for the various plates ; Plate II shows their 
distribution : 



Outside Plates. 

f-inch lap-plate 12 rivets 

f-inch plate 12 " 

^-inch plate 11 " 

^-inch plate 10 " 



Inside Plates. 



j^-inch plate g rivets 

^-inch plate 11 rivets 



Upper Extremity . 

The same thickness of 4I inches of pin-plate is required 
at the upper extremity as at the lower, but a portion of 
this thickness will be formed of an inside diaphragm, partly 
to avoid the inside pin-plates interfering with the web 
members, and partly for the sake of the additional stiffness 
gained by this form of connection. 

The diaphragm will be composed of a tt-inch web, with 
four 4X3iXj inch angles, built in the form of a vertical I. 
Two l^-inch fillers and two ife-inch plates will be added to 
the diaphragm to bear on the pin. Fig. 8 shows clearly 
the dimensions of the pin -plates fastened to the side-plates. 
The computations need not be repeated here, it being^ 
remembered that one half the diaphragm plate thickness 
should be included in the thickness of one jaw. 



End Connection of U1U2. 

The connection of U1U2 at Ui is designed in precisely 
the same manner as the end post connection at Ui. The 
stress in one jaw is 429,000 pounds, and 4.1 inches is the 
thickness of pin-plates required. An inside diaphragm 
will again be used, and Fig. 8 shows clearly the sections 
adopted. 



Art. 22.] 



BENDING OF PINS. 



239 



Since the end post and U1U2 are not riveted together it 
is usual to allow a small clearance between them for ease 
in erection. Fig. 8 shows the clearance adopted. 






' \\1 Ce nter Line 
vJjUZof Truaa 



Center Line of 



Fig. 8. 




It will be seen that open joints are subject to a better 
defined analysis than abutting joints, but they form a 
heavier, more cumbersome, and less rigid connection. 



Art. 22. — Bending of Pins. 

Lower Chord Packing. 

Panel Point L2. 

The packing of pin L2 and the computations of the 
bending moments in it will furnish characteristic opera- 
tions typical of all other cases. Fig. 9 shows the disposi- 
tion of the members meeting at L2, the clearances, and 
distances between centres of adjacent pieces and the 
stresses carried by the pieces when L1L2 has its maximum 



240 



DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VL 



Stress. The maximum stresses in all the members meeting 
at one point are not concurrent, since they are caused 
by different positions of the moving load. It becomes 
necessary to test each pin (i) with the maximum stresses 
existing in the web members, and (2) with the maximum 
stresses existing in the chord members. (It is sufficiently 
accurate to assume that the same position of loading 
causes the maximum stresses in the two adjacent chord 
members or in the web members converging on the pin.) 




CeuterXIce of Truss 



'"^ LsLgltt.OOUlbs. 
=; 1 ' V^ 



« 

Vq 



^Tf- 



-fcjU^T 



3^'X;i'^cld,,5Rl„. 



7"PIN AT L2 

Fig. 9. 



As a rule, the greatest pin-bending in the lower chord occurs 
with the greatest chord stresses, and in the upper chord 
with the greatest web stresses. 

The wind load stresses in the lower chord may be 
neglected as a factor in producing bending in the pins. 

Taking moments about the centre of each piece in Fig. 
9 in turn there is found: 



Art. 22.] 



BENDING OF PINS. 



241 





Horizontal Moment. 


Vertical Moment. 


About b 


+ 181,500 inch-pounds 
+ 151,000 " 
+ 385,000 " 
+ 383,000 
+ 380,000 ' ' 
+ 144,000 " 




c. . 

" d 


330,000 inch-pounds 
330,000 


" / 









The maximura moment, therefore, occurs at e and is 
equal to 



"^383,000 -f 330,000 = 505,000 inch-pounds. 

The allowed unit bending stress is 18,000 pounds per 
square inch, and the bending resistance of a 7 -inch pin 
is 606,100 inch-pounds. The packing at this point is, 
therefore, satisfactory. 

If the bending be computed when L2U1 has its maxi- 
mum stress, the vertical component of which is 166,000 
pounds, it is seen that the vertical moment would be in- 
creased slightly, but the horizontal moment would be more 
than correspondingly decreased. The maximum bending 
moment on the pin occurs in this case, therefore, with the 
maximum chord stress. 

Other panel points in the lower chord are treated in 
precisely the same manner' Plate II shows the packings 
employed. 



Upper Chord Packing. 

Panel Point Uz. 

The packing at panel point Uz may be assumed charac- 
teristic of the entire upper chord; Fig. 10 shows clearly 
the spacing and also the stresses in the different parts, 
when UzMi takes its maximum stress. Since the two 
chord members U2Uz and UzU4, are abutting and as there 



242 



DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 



is practically no eccentricity between them, the resultant 
stress in the web members UzL^ and UzMi, is entirely 
resisted by UzU4„ the horizontal and vertical components 
of which are easily found graphically. 

The maximum stress in UzM^ is +294,000 pounds, and 
the concurrent stress in UzL^ is —86,000 pounds. 

The side-plates of the chord form the supports of the 
ends of the pin, the centre of the supporting forces being 






Eye-bar 7"i l!aa 



^^^TT, 



U,L, 



_I!?Z! 



^m 



"TTM-,. 



C^S" 



\- 



A , (H +107,1 



H-IOY.OOO lbs. 
V-66,000 " 
000 '"' 
000 " 
= 
V 13,000 " 



u„u. 



"3U4 



Fig. 10. 

at the centre of the plates, as shown in Fig. 10. The 
members causing the bending in the pin are UsM^ and 
U3L3, the stress in the latter, -86,000 pounds, being 
wholly vertical. The horizontal and vertical components 
of the stress in one half of UsMi are: 

H. = + 107,000 pounds 
V. = + 99,000 ' ' 

Fig. 10 shows that the bending moment in the central 
portion of the pin is then composed of 

Hor. mom. = 107,000X2^ = 221,000 inch-pounds 

Vert. " = 99,000X21^— 43,000X31 = 48,000 inch-pounds 



Art. 22.] 



BENDING OF PINS. 



'43 



Hence the resultant moment is 



v 221,000^ + 48,000^ = 226,000 inch-pounds. 

The allowed bending resistance of a 6-inch pin is 318,000 
inch-pounds. A 6-inch pin is therefore satisfactory at U3. 

Other upper chord panel points are investigated in 
exactly the same manner. It is desirable to repeat the 
operations in detail for that panel point only at the upper 
extremity of the end posts (Fig. 11), and this point will be 



1-03,500 
1 126,600 




y ^63^600 



Fig. 



investigated for that position of loading which causes the 
maximum stress in U1L2', the following table shows the 
components of the direct stresses in the members meet- 
ing at Ui for that condition of loading: 





Vertical Components. 


Horizontal Components. 


U,L, 


+ 50,000 pounds 
— 620,000 " 
+ 240,000 " 
+ 330,000 " 




LM, 


— i; 10,000 pounds 


U.U, 


+ 790,000 " 


U,Lo 


— 280,000 " 







*44 



DETAILED DESIGN OF A RAILROAD BRIDGE. [Ch. VI. 



No essential error will be introduced if it be assumed 
that there is no eccentricity in the line of action of LqUx 
(the end post) and U1U2. The resulting differences 
between their stresses may therefore be taken exactly 
along their centre lines ; a part of this stress is carried 
by the diaphragm; in this case about 34 per cent. Fig. 11 
shows the stresses carried at one half of the panel point. 

The resultant moments in inch-pounds are as follows : 



Centre of Moments at 
Centre of 



Horizontal Moment 
in Inch-pounds. 



Vertical Moment 
in Inch-pounds. 



Maximum Moment 
in Inch-pounds. 



U,L, 



Centre line of truss. 



<-«« 306,000 
*r-m. 222,500 

:»-> 103,500 



415,000 
345,000 

99,500 



515,000 



The allowed bending resistance of a 7-inch pin is 606,100 
inch-pounds; and this pin therefore satisfies the require- 
ments at Ui. 

Art. 23. — Camber. 

§ loy. All bridges shall he cambered by giving the pands 
of the top chord an excess of length in the proportion of -g- inch 
to every 10 feet. 

In accordance with this specification the following 
table shows the original lengths and the changed lengths 
of the upper chord members. It is seen that the end 
post is not changed in length. 



Original Length. 



Changed Length. 



L,U, 



46' 4" 
30' 6^" 

t t 

29' 61" 
29' 2" 



46' 4" 
3o'^_6f" 

29' ey 

t e 
29' 2j" 



Art. 24.] CONCLUSION. 245 

No changes are made in the lengths of the main web 
members, their final lengths being the same as if no changes 
had been made in the lengths of the upper chord members. 

The stretching of the lower chord panels makes it 
necessary to increase the lengths of MzM^ and M^Mq to 
an amount equal to half that stretch. The final length of 
each of these members will then be 29 feet 2^ inches. 

Art. 24. — Conclusion. 

The principles outlined indicate a system of rational 
treatment for the design of all parts of a modem truss 
bridge. It is obvious that a wide range of treatment of 
details is permissible in securing the above results, and 
those given here are subject to this general observation. 
Other details of equal efficiency might be designed but 
those given are at least as effective as others which might 
have been used and have served the purpose of illustration 
at least as well. 

A carefully made bill of materials with the resulting 
computation of weights would show that the total shipping 
weight of this 350-foot single track through span would 
be 821,000 potmds. 



Plate n. 




INDEX. 



A. 

PAGE 

Anchorage at bases of mill-bents 50 

Arch, masonry, funicular polygon as applied to 18 

, moments in any 133 

, three-hinged, see Three-hinged. 

B. 

Bedplates, design of 223 

Bending, combined, and direct stresses 202 

, of end post 203 

, of supporting columns of roofs 48 

, in pins 229, 239 

Bow 's system of notation 5 

Bracing, stresses in wind, of raikoad bridge 183 

Bridge-truss, deformation of a 168 

, design of a railroad 176 

C. 

Camber 172, 244 

Cantilever, definition of 145 

, moment influence line for 152 

, on towers 1 56 

, reaction influence line for 152 

, shear influence line for 1 50 

, stress influence line for 154 

Character of stress, determination of 7, 10 

Chord members, design of 185, 188, 193 

, influence Une for, any simple truss 96 

, cantilever 1 54 

, three-hinged arch 136 

247 



248 INDEX. 

FAGB 

Chord members, stresses in, truss with parallel chords 77 

web members all inclined 84 

, trusses with subdivided panels 106 

Combined stresses 202 

Composition of displacements 168 

of forces 2 

Contraflexure, point of, in posts 51 

Concurrent forces explained and treated i 

Counterbraces defined 42 

Counter members, design of 200 

, in subdivided panels 103 

Counter-stresses in vertical posts no, 115 

in web members, broken upper chord truss 10 1, 182 

, three-hinged arch 143 

, truss with parallel chords 76 

Cover plate, length of 208, 213 

Crane, stresses in ordinary 12 

Crane-truss, distortion of . 164 

, stresses in 8 

D. 

Dead-load stresses, roof-truss 34 

, simple bridge-truss 48, 178 

, three-hinged arch 128 

, web members, truss with parallel chords 77 

Dead weights of railroad bridge 177 

of roof-trusses 29 

Definition of graphic statics i 

Deflections of structures, see Displacements. 

Deformation of trusses ....,• r6o et seq. 

Design of Parts of Railroad Bridge: 

Bedplates, rollers, and pedestals 223 

Camber 244 

Counter web members 200 

Details at ends of end post 237 

End floor-beams 226 

End post 193 

Floor-beam hangers 194 

Intermediate floor-beams 212 

Lower chord 185 

Lower wind bracing 221 

Main web members 199 

Pin bending 239 

Pin connections of upper chord 234 



INDEX. 249 

PAGE 

Pins and joint details 227 

Portal bracing 220 

Posts. 195 

Stringers 206 

Upper chord 188 

Upper wind bracing 218 

Displacements, composition of two 168 

Displacement diagram, applied to bridge-truss 168 

crane-truss 164 

three-hinged arch 173 

, construction of 161 

Double-intersection trusses, influence lines for 123 

, treatment of 123 

Duchemin's formula for wind components 32 

E. 

End post, bending in 203 

, design of 193 

, design of ends of 237 

Equilibrium of concurrent forces •. 2 

non-concurrent forces 20 

Equilibrium polygon explained 16 

Equivalent uniform load 94 

F. 

Fink roof-truss 42 

Flexure and compression combined 202 

tension combined 203 

Flexure in posts of mill-bents 48 

Floor-beams, design of 212, 226 

Floor-beam hangers, design of '. 194 

Force polygon explained 2 

Funicular polygon, applied to cantilever 146 

, " " three-hinged arch 126 

, explained 16 

, moment polygon 16, 23, 131, 146 

, solution of problems by 20 et seq. 

, special features of 17 

■ , to pass through three points 126 

H. 

Hutton's formula for wind components 32 



2 5° INDEX. 



I. 

PAGE 

Influence areas 69, 94 ' 

Influence line, between adjacent panel points 78 

, definition of 60 

, for counter-stress in vertical post no 

, for double-intersection trusses 123 

, for moments in beams, and simple trusses 67, 69 

in cantilevers 152 

in three-hmged arch 136 

, for reactions of simple trusses 60, 63 

cantilever 147 

three-highed arch 133 

, for shears, simple trusses 61, 66 

, cantifever 150 

, for stresses, in cantilever 154 

, double-intersection truss 122 

, simple truss 91, 117 

, skew bridge 119 

, three-hinged arch 140 

, trusses with subdivided panels 10 j 

J- 

Joint details, design of 228 

K. 

King-post truss, stresses in 7 

Knee-brackets, stresses in 48 

L. 

Live-load stresses in railroad bridge, chord members 180 

, web members 181 

Lower chord members, design of 185 

Lower chord wind bracing 221 

M. 

Method of moments, bridge-truss 179 

, Fink truss 44 

Mill-bent, stresses in 48 

Moment, by funicular polygon 16 

, criterion for maximum, simple beam 69 

, Warren truss 80 

, in a beam 23 



INDEX. 2 SI 



PAGE 



Moment in a cantilever 1 46 

in three-hinged arch 131 

influence hne for beam or simple truss 67 

for cantilever 152 

for three-hinged arch 136 

for Warren truss 79 

maximum value of, in a beam 69 

of inertia of posts 1 96 

of upper chord members 191 



N. 

Non-concurrent forces, force polygon for 13 

members, stresses in three 28, 43, 87 

Notation, system of 5 

P. 

Panel loads for roof-trusses 32 

Parallel forces, moment polygon for 23 

Pedestals, design of 223 

Pin connections of upper chord, design of 234 

plates in posts 231 

Pins, bending of 239 

, design of 227 

Pitch of roofs 30 

Pole distance 16 

Pole explained 16 

Portal bracing, design of 220 

Posts, design of 195 

, counter stresses in no, 115 

Purlins 32 

R. 

Radius of gyration of posts 196 

of upper chord member 192 

Rays 16 

Reaction influence lines, for beam, single load 60 

, series of loads 63 

, cantilever 147 

on tower 158 

, three-hinged arch 133 

Resolution of concurrent forces i 

Riveting in flanges of girder 209, 214 



252 INDEX. 

PAGB 

Rollers, design of, for bridge 223 

, for roof-trusses 30 

Roof covering, weights for 32 

Roof-trusses, dead weight of 30 

, snow load for 30 

, stresses in 29 et seq. 

, wind loads for 31 

S. 

Scales for influence lines 93, 96, 138 

Shear influence line for beam and simple truss 61, 66 

for cantilever 150 

Shear, maximum, in truss with parallel chords 74 

, variation of, within a panel 75 

Skew bridges 119 

, influence lines for 119 

Snow loads for roof-trusses 30 

stresses for roof-trusses 36 

Specifications for design of railroad bridge, throughout Chapter VI. 

Spandrel-braced arch 128 

Stresses, chord, in railroad bridge 180 

, in roof-truss, both ends fastened 33 

, one end on rollers 38 

, influence Unes for, cantilever 154 

, simple bridge, trusses 74- et seq 

, three-hinged arch 136, 140 

, web 181 

, wind, in railroad bridge 184 

Stringers, design of 206 

Subdivided panels, treatment of 103 

Substitution of diagonals (Fink truss) 45 

Suspension-bridge cable, funicular polygon as 20 

T. 

Three-hinged arch 126 

, deflection of 173 

, influence line for chord members 136 

reactions 133 

web members 140 

, moments in 131 

, reactions of 128 

Thrust, in three-hinged arch 130 

, maximum, in three-hinged arch 135 

Tower, cantilever on 156 



INDEX. 253 



PAGE 

Unsymmetrical trusses 46 

Upper chord members, design of 1 88 

lateral wind bracing, design of 218 

, stresses in 183 

V 

Variation of influence line within a panel 78 

moments within a panel 83 

W. 

Web member, counter, design of 200 

, criterion for maximum stress, simple truss 97 

, influence line, cantilever 

, simple truss 91 

, three-hinged arch 140 

, main, design of 199 

, stress in, any simple truss 88, 117 

, double-intersection truss 124 

, skew bridge 119 

, truss with parallel chord 74 

subdivided panels 105 

Weight of railroad bridge 178 

of roof covering for roof-trusses 32 

of roof -truss 30 

Williot diagram, see Displacement. 

Wind bracing, design of lower 221 

upper 218 

, specifications for 217 

Wind loads for roof-trusses 31 

Wind stresses in railroad bridge 183 



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